$.99$ Store

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.

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

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

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

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

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

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.

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

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

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

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

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

Meaning, an integer like $53 = 5.10^1 + 3.1$ . 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 $11$ . For a three digit number, the number of such quantities would therefore be $111$ and so forth.

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

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

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

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

Hence the answer is 11.

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

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

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

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

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

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.

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

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

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.

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

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

(+)Left/(-)over: 0

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

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".

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🤷‍♂️

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).

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.

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.

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}

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

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 ;-)

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

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

1089/99=11