. 99 .99 Store

Algebra Level 1

Marie shops at a store where all prices end in 99 cents ($0.99, $1.99, $2.99, etc.). She ends up spending $33.89.

How many items did she purchase?

8 9 10 11 It's impossible to know for sure

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31 solutions

For every item she buys, the price she pays is $0.01 short of a round number. By paying $33.89, the total price paid is $0.11 short of a round number. She then bought 11 items.

Exactly - short and succinct. For completeness one can point that 111 items (the next possibility up) would cost a minimum of $109.89 (111 x 0.99) and is therefore ruled out.

Thomas Sutcliffe - 2 years, 9 months ago

cannot be since the highest price 2.99. If she buy 11 of them the total only 32.89 not 33.89

Pappagallo Villas - 2 years, 9 months ago

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I agree. why is the answer 11, it can be 12 ( buying another 99 cent item. Can someone explain this anomaly?

fynx gloire - 2 years, 9 months ago

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That would make the decimals 88, i.e. 1 cent short of the actual amount spent.

The idea is that all products are priced a + $ 0.99 = ( a + $ 1 ) $ 0.01 a +\$0.99 = (a+\$1) - \$0.01 with a a a whole number of dollars. The total of n n of these products, therefore is is a whole number, minus n n cents. As the total spent is 11 cents shy of a whole number, n = 11 n=11 (or 111, 211, etc.), but it is easy to demonstrate that these higher numbers are impossible, because much more money would have been spent.

Roland van Vliembergen - 2 years, 9 months ago

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@Roland van Vliembergen Hello, thx for your reply, so you are saying: the sum of 1 .. n of (a+$1.00)-$0.01 which equals a whole number plus .89 cents.

Your quote of: "The total of n of these products, therefore is is a whole number, minus n cents"

How do you get its a whole number minus n cents from? Can you continue with your formula to explain that statement?

regards

fynx gloire - 2 years, 9 months ago

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@Fynx Gloire The value of a a can change based on the price of the product, so when we buy n n products, we should label the values a 1 , a 2 , . . . a n a_1, a_2, ... a_n . Then the total price based on the formula is ( a 1 + $ 1 ) $ 0.01 + ( a 2 + $ 1 ) $ 0.01 + . . . + ( a n + $ 1 ) $ 0.01 = ( a 1 + $ 1 ) + ( a 2 + $ 1 ) + . . . + ( a n + $ 1 ) $ 0.01 $ 0.01 . . . $ 0.01 = ( a 1 + $ 1 ) + ( a 2 + $ 1 ) + . . . + ( a n + $ 1 ) $ 0.01 × n (a_1+\$1)-\$0.01+(a_2+\$1)-\$0.01+...+(a_n+\$1)-\$0.01 = (a_1+\$1)+(a_2+\$1)+...+(a_n+\$1)-\$0.01-\$0.01-...-\$0.01 = (a_1+\$1)+(a_2+\$1)+...+(a_n+\$1)-\$0.01\times n . The first part has to be an integer because the values in parentheses are all integers.

The number of cents is decided by the part at the end, which subtracts one cent each time a new product is bought. Since the number of cents is 89 89 in this problem, we have to subtract 11 11 cents from a dollar. So n = 11 n=11 .

Zain Majumder - 2 years, 9 months ago

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@Zain Majumder ahh ok I think I see what you are doing. So basically in your formula (a1+$1)+...+(an+$1) HAS TO EQUAL 34 since its the next integer greater than 33.89 correct? Then we just solve for n where, 34 - 0.01 x n = 33.89, hence n = 11?

fynx gloire - 2 years, 9 months ago

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@Fynx Gloire Yep. That's correct :)

Zain Majumder - 2 years, 9 months ago

Maybe the problem has changed, but it doesn't say the highest price is 2.99 -- just that every item ends in .99. It could be that the first item is 23.99 and there are ten other 0.99 items.

Greg Whiteside - 2 years, 9 months ago

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Indeed if you look closely the question makes it qjuite clear that there ARE items priced above £2.99 "...($0.99, $1.99, $2.99, etc.)" note the etc., short for etcetera, which means that are items in the store priced at $3.99, $4.99 and other such prices.

Thomas Sutcliffe - 2 years, 9 months ago

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@Thomas Sutcliffe Further to the above, there is also a pair of scissors priced at $3.99 shown in the picture, rendering Mr Villas' objection utterly spurious.

Thomas Sutcliffe - 2 years, 9 months ago

The question never stated a price range, the only rule was the price ends with .99c price.

This is why over 40% of people got this question wrong, lol.

Ephraim Koay - 2 years, 9 months ago

En la pregunta esta la respuesta dice precios desde . 99, 1.99,2.99,3.99 etc significa que puede haber cosas de 4.99, 5.99,6.99 etc. Entonces compro 4 de .99, 1 de 1.99, 2 de 2.99, 1 de 3.99, 1 de 4.99, 1 de 5.99 y otra de 6.99 dando una respuesta la respuesta de 33.89

julio granillo - 2 years, 9 months ago

I'm surprised 40% of people got this wrong...

Simon The Great - 2 years, 9 months ago

What if Marie bought 1 item of 31.99 and another item of 1.99. So I don’t think 11 is the only solution

Francisco Flores - 2 years, 9 months ago

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That would be $33.98, not $33.89

Zain Majumder - 2 years, 9 months ago
Stephen Mellor
Aug 27, 2018

1 item means that the cent value of the total will be 99 \text{1 item means that the cent value of the total will be } \boxed{99} 2 items means that the cent value of the total will be 99 + 99 98 (mod 100) \text{2 items means that the cent value of the total will be } 99 + 99 \equiv \boxed{98} \text{(mod 100)} \vdots n items means that the cent value of the total will be 100 n n \text{ items means that the cent value of the total will be } \boxed{100 - n} 100 n = 89 n = 11 \implies 100 - n = 89 \implies \boxed{\color{#D61F06}{n=11}} (The next possibility would be 111 items to get the correct pence, but 111 × 0.99 is the minimum which is way more than 33.89 ) \text{(The next possibility would be 111 items to get the correct pence, but } 111 \times 0.99 \text{ is the minimum which is way more than 33.89})

what is mod?

Gilang AB - 2 years, 9 months ago

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See this wiki page

Stephen Mellor - 2 years, 9 months ago

Let the number of items Marie purchased from the store by n n . The maximum possible value of n 33.89 0.99 = 34 n \le \left \lfloor \dfrac {33.89}{0.99} \right \rfloor = \color{#D61F06}34 . Using cents instead of dollars, we have:

99 n 89 (mod 100) ( 100 1 ) n 89 (mod 100) n 89 (mod 100) n 89 11 (mod 100) \begin{aligned} 99n & \equiv 89 \text{ (mod 100)} \\ (100-1)n & \equiv 89 \text{ (mod 100)} \\ -n & \equiv 89 \text{ (mod 100)} \\ \implies n & \equiv -89 \equiv 11 \text{ (mod 100)} \end{aligned}

The next possible n n is 111 > 34 111 >\color{#D61F06} 34 , hence n = 11 n=\boxed{11} is the only solution.

How do you get ( 111 > 34 ) (111>34) ?

Gia Hoàng Phạm - 2 years, 9 months ago

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See my second sentence that established that with $33.89, Marie could only buy the most 34 items.

Chew-Seong Cheong - 2 years, 9 months ago

How about 111?

Gia Hoàng Phạm - 2 years, 9 months ago

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n 11 (mod 100) n \equiv 11 \text{ (mod 100)} means n m o d 100 = 11 n \bmod 100 = 11 . That is, when n n is divided by 100, the remainder is 11. Of course when 11 is divided by 100, the remainder is 11. The next number is 111 and then 211, 311, 411...

Chew-Seong Cheong - 2 years, 9 months ago

it's possible to have bought only 2 items...one is 32.99 the other is .99 or one is 31.99 and the other is 1.99 so there's no way to know

Margaret Longo - 2 years, 9 months ago

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Then it would be 32.99 + 0.99 = 33.98 33.89 32.99+0.99 = 33.98 \ne 33.89 .

Chew-Seong Cheong - 2 years, 9 months ago

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ooops. thanks. I read too quickly!

Margaret Longo - 2 years, 9 months ago

Among these four answer only 11 Time 9 give us last desimal no 9 =>answer is 9

Mayank Bhandari - 2 years, 9 months ago

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But first decimal will be zero rather than 89.

Nagesh S - 2 years, 9 months ago

She starts from 0.99 and every item she buys decreases 0.01 so the answer is 10 why 11??

Dan Gal - 2 years, 9 months ago

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1 0.89 = 0.11 = 0.01 × 11 1-0.89=0.11 = 0.01 \times 11

Chew-Seong Cheong - 2 years, 9 months ago

remember.. 11 x 99 = 10.89

J . - 2 years, 9 months ago
Gghh Tty
Sep 1, 2018

If you take the number 99 and sum 99 ten times, you get a number ended in 89

99 x 10 = 990, not 989

Steven Adler - 11 months, 2 weeks ago

99 * 11(%100)=89(%100).

So, the number of items she buys has last 2 digits as 11.

11 * .99=10.89

111 * .99=109.89

So if she buys 111 items or more, her total cost would have been more than 33.89.

Hence, she buys 11 items.

The last digit of 33.89 is 9 and the last digit of every price item is 9.Having that the posible quantities could be 1,11,21,31 ...the quantity cannot be one because 33.89 is lower than 1x33.99(the nearest price). 8,9,10 are excluded because last digit cannot be 9.It means that quantity can be 11,but we still have to demostrate that. If quantity is 11 then we have decimal part equal to 11x0.99=10.89. From 33.89 we cut the 9.89 and it remains 23. Can we combine (sum) 11 integer numbers to create the number 23? Yes we can . 23 = 2x10 + 1×3 That means 10 pieces from product with price 2.99 and 1 piece from product with price 3.99.

Antoneta Prifti - 2 years, 9 months ago
Carlo Carparelli
Sep 4, 2018

Every items as a cost of .99 Total costs is *.89 The only way to obtain 89 from 99 is 11, infact: 8x99= *.92 - 9×99= .91 - 10×99=*.90

Shaurya Singh
Sep 5, 2018

We can start by breaking each individual price into x i 0.01 x_i - 0.01 , where x i Z + x_i \in \mathbb{Z}^+ and define the following relation

i = 1 k i x i k ( 0.01 ) = 33.89 \sum\limits_{i=1}^{k}ix_i - k(0.01) = 33.89

Any n n digit number having d i d_i digit on i t h i^{th} place can have a decomposition

d = i = 1 n d i ( 1 0 i 1 ) d = \sum\limits_{i=1}^nd_i(10^{i-1}) ..... (1)

Meaning, an integer like 53 = 5.1 0 1 + 3.1 53 = 5.10^1 + 3.1 . Here we can think of the digits as the price and the powers of ten as the quantity, which in this case would sum to 11 11 . For a three digit number, the number of such quantities would therefore be 111 111 and so forth.

We can put, k = 11 k=11 for our equation and get

i = 1 11 i x i 11 ( 0.01 ) = 33.89 \sum\limits_{i=1}^{11}ix_i - 11(0.01) = 33.89

i = 1 11 i x i = 34 \sum\limits_{i=1}^{11}ix_i = 34

This conforms to (1) where we need 10 10 of one price and 1 1 of another and the rest are 0 0 .

Hence the answer is 11.

Darth Korma
Sep 3, 2018

For every item that is bought, a cent is to be deducted from the total e.g. if 47 items are bought for $0.99, then $0.47 is to be taken away from $47.00:

$47.00 - $0.47 = $46.53

J .
Sep 7, 2018

Consider this. everything ends in 99, so 11 × 9 = (9)9. 10 x 9 = (9)0. Can't be that.. 9 x 8 = (7)2.. nope, ends in 2, not 9. Therefore, the answer has to either be 11 or impossible to discern. .99 x 11 is.. you guessed it! 10.89. Now all you must do is account for the other $23. Ergo, the correct answer is "11". please notr it would not be possible to determine how much was paid for each item. Why? One. possible solution would be 10 items at $2.99 and one @ $4.99. There are several possible solutions.. try it and see what I mean. It's not going to matter as long as your last two numbers are 89

Juan Guzman
Sep 6, 2018

I saw 33.89 so I though we'll If every item is 0.99 then 2 items would be 1.98 for 3 item 2.97. I noticed the last digit kept decreasing, so I multiplied 0.09 until it ended in 0.09 again.

{0.99 x 2 = 1.98} ; {0.99 x 3 = 2.97} ;... {0.99 x 10 = 9.90} ; {0.99 x 11 = 10.89} ;

It took me 11 items to get 00.89 to match 33.89

Nour Jamal
Sep 3, 2018

Well I figured it out in a kind of a weird but simple way but not 100% correct If u add the number of prices of the items 0.99+1.99+2.99+3.99=9.96.

She paid 33.89 ,

if u divide 33.89 by the sum u will get 3.4026 or simple say 3 ,(note i now replace 3.4026 by 3 in the rest of this comment)

this means if u multiply the sum of the items by 3 then u will get the amount she paid (9.96)3 or (0.99+1.99+2.99+3.99)3 , to make this simpler if u use the distributive property this means u bought at least three pieces of each item. Since we have 4 items, and u bought 3 pieces of each of these items or 3*4 = 12 (total number of items), Since it's a multiple choice question, u can choose the highest number (11) items

Vinod Kumar
Sep 2, 2018

3289 has prime factors 11, 13, 23 with 299 is a divisor, thus giving 3289/299=11. Answer=11

33.89 spent (3389)

Roland Kensdale - 2 years, 9 months ago

3389 is a prime

Guy Fox - 2 years, 9 months ago
Dave Ames
Sep 9, 2018

The 33 dollars is a distraction. How do you end up with 89 cents? The only way to achieve that number is by adding 99 cents to a number ending in .90 (or, in this case 32.90 and 99 cents). Therefore, we know that there are at least 10 items by .99 * 10 = 9.90. So X must be 11, or 10 items totaling A.90 and an additional item for .99 to equal B.89.

Ephraim Koay
Sep 7, 2018

Every time you add .99, the last two values will be 0.01 less in value.

100 - 11 = 89 , therefore, 11 .

I don't know why people do complicated solutions to this... The cents is .89 , and .99 is 1-.01 . since it's .89, it's (rounded price of objects she bought) -.11 . She then bought 11 objects which prices are unknown except for the cents. Just, why all the equations? lol

You think my solution was a complicated equation? I simply created a simple theorem and proved it, using simple deductive logic. However, if you think about it, isn't it nice that so many people can take so many 'different roads' to get the same place? (Thereby proving each other's results!)

J . - 2 years, 9 months ago
Calvin Kong
Sep 6, 2018

(0.99 x 3) + (1.99 x3) +2.99+3.99+4.99+5.99+6.99 =33.89

10 2.99+1 3.99=33.89 LOL

Guy Fox - 2 years, 9 months ago

A not so elegant and brute force way... maybe someone can help share the logic/equation for this. Step 1. Start by buying the most expensive item that can be bought (3.99) for the smallest quantity (8) in the choice list. Step 2. Calculate how much money left. Step 3. Determine the next expensive item that can be bought at maximum quantity with the money left. Step 4. Keep repeating Step 3 until you cannot buy anything else. Step 5. If there's no money left, we are done. If there's still money left, reduce the number of the most expensive item in the list and repeat step 3 - 4.

Price Qty Sub-total
3.99 8 31.92
2.99 0 0
1.99 0 0
0.99 2 1.98

(+)Left/(-)over: -0.01

Price Qty Sub-total
3.99 7 27.93
2.99 0 0
1.99 3 5.97
0.99 2 0

(+)Left/(-)over: -0.01

Price Qty Sub-total
3.99 5 19.95
2.99 3 8.97
1.99 2 3.98
0.99 1 0.99

(+)Left/(-)over: 0

N Kansara
Sep 3, 2018

Here all the items have price x + 0.99 where x is a whole Number. Since the total price is 33.89 , the number of items must contain 11 as last 2 digits(because 99× (k×100 + 11) = (n×100 +89 ) . If the no. Of items = 11 the we can somehow sum it up to 33.89. BUT if the number of items is a three digit no. Of form k11 where k is a natural no. then the sum will always exceed 109.89 which we don't want. Thus 11 is the only solution

--- ---
Jun 14, 2020

Yes, for every item that she buys, her "cent count" is decreased by one. Because the lowest priced option is $0.99 she couldn't have been able to buy 111 items in order to satisfied the problem the same way as well. So, there is only one answer. Her cent count was $0.89, which is 11 away from the cent count "$0.00".

Hriday Gupta
Jan 1, 2019

forget everything and assume that evry object had been priced at $1,2,3,or 4 now we have increased the price of every object by 1 cent it must be compensated in our price hence our paid price increases by n cents for the n objects we buy

THE ADVANTAGE: since all our new prices are integers our final price must also be a integer thus 0.01*n+39.89 must yield an integer! 11 being the answer

I basically just divided 33.89 by 2.99, assuming everything he bought was $2.99. I got an approximate value to 11. Thats what i did. Not the best solution but🤷‍♂️

Joel Vaz
Sep 9, 2018

Check for the table of 9.so everything she buys will end up in 9 x (the number of items which will probably end in xxxx.99).

Sadiq Musanna
Sep 9, 2018

My approach was simple. I checked every option and multiply it with 0.99 and for 11 it matches the total amount spend as Its fractinal value was. 89.

Gourav Ganguly
Sep 8, 2018

I look at the options available- multiplying by 8 creates a 2 at the end of the result, 9 creates a 1, 10 a zero and so on. But 11 has a 1 and multiplying it will create a 9 and this matches the answer. So no, it’s not an impossible guess.

Ritwik Rudra
Sep 8, 2018

Since each [price 100] ends with 9 as unit place, we can say that for getting the [sum 100] to be ending with 9, we have only 11 as probable answer.{Since 9 *8,9 9,9 10 all do not end with 9: but 9 11=99 ends with 9}

further 0.99*11=10.89

33.89-10.89= 23

so we need to have sum of eleven natural numbers to be equal to 23. this is very much possible{ one such case is TEN 0.99 & ONE 23.99}

Hael Uresi
Sep 8, 2018

The last two digits are the cents. 0.99*11=10.89 so the last 2 are 89 and the total she used was 33.89 this shows that she bought 11 items

Brent Mullins
Sep 7, 2018

The question should specify that it is a store where all prices end in 99 cents and sales tax has already been included in the price . . . you know, for us Americans ;-)

Kushagra Tiwari
Sep 7, 2018

Simple solution Adding 9 'n' times gives '9' in the unit place only if 'n' has 1 in its unit place. Eg: 9+9= 18 9+9+9=27 and so 11 times 9=99 21 times 9=189 Therefore, by just looking at options we can find the answer, as only the option (d) has 1 at unit's place.

Used calculator

Leung Hong
Sep 4, 2018

Just add 0.99 + 0.99 until you get .89 at the back. I pressed for 11 times

Ervyn Manuyag
Sep 2, 2018

1089/99=11

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