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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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 $a +\$0.99 = (a+\$1) - \$0.01$ with $a$ a whole number of dollars. The total of $n$ of these products, therefore is is a whole number, minus $n$ cents. As the total spent is 11 cents shy of a whole number, $n=11$ (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 $a$ can change based on the price of the product, so when we buy $n$ products, we should label the values $a_1, a_2, ... a_n$ . Then the total price based on the formula is $(a_1+\$1)-\$0.01+(a_2+\$1)-\$0.01+...+(a_n+\$1)-\$0.01 = (a_1+\$1)+(a_2+\$1)+...+(a_n+\$1)-\$0.01-\$0.01-...-\$0.01 = (a_1+\$1)+(a_2+\$1)+...+(a_n+\$1)-\$0.01\times n$ . 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 $89$ in this problem, we have to subtract $11$ cents from a dollar. So $n=11$ .

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

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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what is mod?

Gilang AB
- 2 years, 9 months ago

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.

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

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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$n \equiv 11 \text{ (mod 100)}$ means $n \bmod 100 = 11$ . That is, when $n$ 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 $32.99+0.99 = 33.98 \ne 33.89$ .

Chew-Seong Cheong
- 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

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

remember.. 11 x 99 = 10.89

J .
- 2 years, 9 months ago

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

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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.

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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

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

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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.

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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

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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

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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

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

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33.89 spent (3389)

Roland Kensdale
- 2 years, 9 months ago

3389 is a prime

Guy Fox
- 2 years, 9 months ago

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

100 - 11 =
**
89
**
, therefore,
**
11
**
.

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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

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

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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

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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

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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}

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

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improve more

improve more
- 2 years, 9 months ago

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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.