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If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha

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If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 15 May 2019, 04:05
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A
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C
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Question Stats:

40% (02:42) correct 60% (02:11) wrong based on 30 sessions

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Re: If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 15 May 2019, 11:26
My observation:
Since 19! is divisible by numbers 1-19 because it's essentially the product of it, we know that that 18 is one of those numbers.
If a number is divisible by 18, the sums of digits should be divisible by 9.
Hence:
33 + (T+H+M) should be divisible by 8
33 + 3 is 36
So T+M+H becomes 3.

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Re: If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 15 May 2019, 12:01
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Extremely good question. Let me give it a try.

First and foremost, we will try to identify the values of all the variables: T,M, and H

Firstly, the number is 19!. It has 5 appearing 3 times (5,10,15). So, it must end with three zeroes at the end.

Hence, the value of H must be 0.

Now, because 19! is 1*2*3*......*18*19, it is definitely divisible by 9.
Test of divisibility with 9 states that the sum of the digits must be a multiple of 9.
Thus, sum of the digits of 19! is 33 + T+M+H).
As, H = 0, we need 33 + (T+M) to be a multiple of 9.
So, T+M can be either 3 or 12 but nothing beyond it as maximum addition of T and M is 18.

Case 1: T+M = 3
Case 2: T+M = 12

Also, 19! is divisible by 11.
According to test of divisibility with 11, the difference between the sum of digits in odd places and sum of digits in even places must be a multiple of 11.
So,
Sum of digits at even places = 20+M
Sum of digits at odd places = 13+T

Case 3: 20+M - (13+T) = 7+M-T
M-T = 4 or M-T = -7
Now,
T+M = 3 and M-T = -7 does not lead to any solution for T and M
Also, T+M = 3 and M-T = 4 does not lead to any solution for T and M.

We can solve for T+M = 12 and M-T = 4

BUT

AS THERE IS ONLY ONE POSSIBILITY FOR T+M and Value of H is also known,
T+M+H = 12

Option C is our answer.

No need to check for the scenarios where T-M = 4 or T-M = -7 as again it will lead us to T+M = 12 as the only possible case.


If you like my explanation, hit the Kudos and feel free to ping me for any clarification.

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If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 15 May 2019, 17:32
Hi vinit, How did you get, M-T=4.

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Re: If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 15 May 2019, 19:34
Hi Raxit85

I applied the test of 11 which leads to 7+M-T. Now 7+M-T must be a multiple of 11 which leads to M-T to be 4.

Does that clear your doubt?

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Re: If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 15 May 2019, 20:24
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Bunuel wrote:
If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits that are not given. What is T+M+H?

(A) 3
(B) 8
(C) 12
(D) 14
(E) 17


19! has three 5s so it will end with three 0s.
Hence H = 0

19! = 121,6T5,100,40M,832,000 = 121,6T5,100,40M,832 * 1000

19! has many 2s. So M832 will be divisible by 16 (last 4 digits).

So M000 + 832 is divisible by 16. 832 is completely divisible by 16. Since 16*5 = 80, 8000 must be divisible by 16.
M = 8

19! = 121,6T5,100,408,832,000
Since 19! is divisible by 9, sum of all digits must be divisible by 9.

1+2+1+6+T+5+1+4+8+8+3+2 = T + 41

To make this divisible by 9, T must be 4.

H+M+T = 0 + 8 + 4 = 12

Answer (C)
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Re: If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 18 May 2019, 00:13
VeritasKarishma wrote:
Bunuel wrote:
If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits that are not given. What is T+M+H?

(A) 3
(B) 8
(C) 12
(D) 14
(E) 17


19! has three 5s so it will end with three 0s.
Hence H = 0

19! = 121,6T5,100,40M,832,000 = 121,6T5,100,40M,832 * 1000

19! has many 2s. So M832 will be divisible by 16 (last 4 digits).

So M000 + 832 is divisible by 16. 832 is completely divisible by 16. Since 16*5 = 80, 8000 must be divisible by 16.
M = 8

19! = 121,6T5,100,408,832,000
Since 19! is divisible by 9, sum of all digits must be divisible by 9.

1+2+1+6+T+5+1+4+8+8+3+2 = T + 41

To make this divisible by 9, T must be 4.

H+M+T = 0 + 8 + 4 = 12

Answer (C)


in So M000 + 832 is divisible by 16. 832 is completely divisible by 16. Since 16*5 = 80, 8000 must be divisible by 16.
M = 8
M can also be 2 or 4 right??
why are we not considering those?
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If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 18 May 2019, 01:28
smita123

You are absolutely correct. in fact, M can be 2/4/6/8/0 also.

That’s why you will have to look at other conditions too.

Please go through my solution provided above and let me know if you have some doubt.

VeritasKarishma

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If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 18 May 2019, 02:47
smita123 wrote:
VeritasKarishma wrote:
Bunuel wrote:
If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits that are not given. What is T+M+H?

(A) 3
(B) 8
(C) 12
(D) 14
(E) 17


19! has three 5s so it will end with three 0s.
Hence H = 0

19! = 121,6T5,100,40M,832,000 = 121,6T5,100,40M,832 * 1000

19! has many 2s. So M832 will be divisible by 16 (last 4 digits).

So M000 + 832 is divisible by 16. 832 is completely divisible by 16. Since 16*5 = 80, 8000 must be divisible by 16.
M = 8

19! = 121,6T5,100,408,832,000
Since 19! is divisible by 9, sum of all digits must be divisible by 9.

1+2+1+6+T+5+1+4+8+8+3+2 = T + 41

To make this divisible by 9, T must be 4.

H+M+T = 0 + 8 + 4 = 12

Answer (C)


in So M000 + 832 is divisible by 16. 832 is completely divisible by 16. Since 16*5 = 80, 8000 must be divisible by 16.
M = 8
M can also be 2 or 4 right??
why are we not considering those?


This is a GMAT PS question. I will have only one correct answer. The question says "What is T+M+H?"
In all cases, T+M+H should take the same value so as long as I have one set of values that satisfies, I will ignore others.
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If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha  [#permalink]

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New post 18 May 2019, 03:09
VeritasKarishma wrote:
Bunuel wrote:
If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits that are not given. What is T+M+H?

(A) 3
(B) 8
(C) 12
(D) 14
(E) 17


19! has three 5s so it will end with three 0s.
Hence H = 0

19! = 121,6T5,100,40M,832,000 = 121,6T5,100,40M,832 * 1000

19! has many 2s. So M832 will be divisible by 16 (last 4 digits).

So M000 + 832 is divisible by 16. 832 is completely divisible by 16. Since 16*5 = 80, 8000 must be divisible by 16.
M = 8

19! = 121,6T5,100,408,832,000
Since 19! is divisible by 9, sum of all digits must be divisible by 9.

1+2+1+6+T+5+1+4+8+8+3+2 = T + 41

To make this divisible by 9, T must be 4.

H+M+T = 0 + 8 + 4 = 12

Answer (C)


Hi VeritasKarishma

M000 can be 2000 also right, 2000 being divisible by 16. In this case M becomes 2 and T becomes 1, giving H+M+T=3

Please clarify
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If 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits tha   [#permalink] 18 May 2019, 03:09
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