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Bunuel
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 11 Jan 2022, 08:45
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49% (02:52) correct 51% (02:30) wrong based on 164 sessions

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In the quotient \(\frac{25!}{15!}=11,861,676,2AB,000\), what is the product of the digits represented by A and B?

(A) 64
(B) 66
(C) 68
(D) 70
(E) 72


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A
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bacca2323
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 12 Jan 2022, 01:10
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I'm not sure put I will try to put it in this way: (25!) /(15!) = 25 x 24 x 23 ... x 18 x 17 x 16. So it must be divisible by 9 since 18 has 9 as a factor. Therefore the digit must add for a multiple of 9. We can see that excluding A and B the total sum is equal to 38, so AB must add for at least 7 or 16.

7 can be obtained by adding 4 and 3, 6 and 1, 7 and 0 or 5 and 2, but the product between these options is not in the answers. So they must add for 16, which can be made with 9+7 or 8+8. 9x7 = 63 which is not in the option, while 8x8 equals 64, which is letter A.

A is the answer IMO
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GathaPrajapati
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 15 Dec 2023, 00:08
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AB should be single integer individually. Anything above 64 cannot be expressed as factors of digits less than 9. I'd by trial and error 8x8=64, 6x11 eliminate and so on...
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 16 Oct 2024, 02:42
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You can find unit B by eliminating the 3 5s and 3 2s from the equations that would make the the 0s and finding the units digit that is 8... then you can say that the number is a multiple of 9 and add the digits to find A.. 8x8=64
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 17 Oct 2024, 23:42
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Bunuel
In the quotient \(\frac{25!}{15!}=11,861,676,2AB,000\), what is the product of the digits represented by A and B?

(A) 64
(B) 66
(C) 68
(D) 70
(E) 72


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(25!) /(15!) = 25 x 24 x 23 ... x 18 x 17 x 16

25 x 22 x 20 = 11000 -> which justifies the 3 zeros towards the end; then just multiply all the remaining digit's unit, it's last 2 digits will be 8 and 8.

Unsure of the method, but that's how I got my answer. Can someone please ratify the same.
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 18 Oct 2024, 00:08
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Anurag18
Bunuel
In the quotient \(\frac{25!}{15!}=11,861,676,2AB,000\), what is the product of the digits represented by A and B?

(A) 64
(B) 66
(C) 68
(D) 70
(E) 72


Are You Up For the Challenge: 700 Level Questions
(25!) /(15!) = 25 x 24 x 23 ... x 18 x 17 x 16

25 x 22 x 20 = 11000 -> which justifies the 3 zeros towards the end; then just multiply all the remaining digit's unit, it's last 2 digits will be 8 and 8.

Unsure of the method, but that's how I got my answer. Can someone please ratify the same.
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The important thing is not only that 25 x 22 x 20 = 11,000 ends in 3 trailing zeros, but also that the numbers before that are ones. Additionally, it's not entirely clear how it's so easy to determine the last two digits of the product of the units digits.
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 18 Oct 2024, 00:35
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In the quotient \(\frac{25!}{15!}=11,861,676,2AB,000\), what is the product of the digits represented by A and B?

\(\frac{25!}{15!}= 25*20*19*18*17*16 = 11,861,676,288,000\)

A = 8; B=8; A*B = 8*8 = 64

IMO A
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 18 Oct 2024, 21:36
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Kinshook
In the quotient \(\frac{25!}{15!}=11,861,676,2AB,000\), what is the product of the digits represented by A and B?

\(\frac{25!}{15!}= 25*20*19*18*17*16 = 11,861,676,288,000\)

A = 8; B=8; A*B = 8*8 = 64

IMO A
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Kinshook I do not understand your explanation.
I think you should provide much details of this, how you got your answer. Otherwise anyone can write these kind of explanation.
Bunuel please check this!!
Bunuel please provide the official solution.
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 22 Oct 2024, 04:32
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Is there a solution?
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kayarat600
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In the quotient 25!/15! = 11,861,676,2AB,000, what is the product of 22 Oct 2024, 04:35
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I kind of understand the answer here but still have doubts about how the seperatoojnof 11,000 worked also the Calc of this method is too time confusing
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