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15 May 2019, 04:05
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (02:42) correct
60%
(02:27)
wrong
based on 47
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 3
(B) 8
(C) 12
(D) 14
(E) 17
Vinit800HBS
Expert
Tutor
Joined: 29 Dec 2018
Last visit: 22 Mar 2026
Posts: 90
Given Kudos: 195
Location: India
Schools: HBS '22 Wharton '22 Stanford '22
GRE 1: Q170 V163
Expert reply
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.
Posted from my mobile device
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.
Posted from my mobile device
General Discussion
15 May 2019, 11:26
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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.
Posted from my mobile device
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.
Posted from my mobile device
