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If n is a 6-digit positive integer, abcd96, where a is the hundred-thousands digit of n, b is the ten-thousands digit of n, c is the thousands digit of n, and d is the hundreds digit of n, is n divisible by 36?
(1) The remainder is 3 when \(a+b+c+d\) is divided by 9.
(2) \(a+b+c+d\) is divisible by 3.
Are You Up For the Challenge: 700 Level Questions
(1) The remainder is 3 when \(a+b+c+d\) is divided by 9.
(2) \(a+b+c+d\) is divisible by 3.
Are You Up For the Challenge: 700 Level Questions
18 Jan 2021, 01:28
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The Integer N will be divisible by 36 if it is divisible by the Prime Bases that make up 36’s Prime Factorization.
36 = (2)^2 * (3)^2
Divisibility Rule for (2)^n power
If the last n digits are divisible by (2)^n
Then the number will be divisible by (2)^n
As given, the last 2 digits of n are 96, which is divisible by 4
Thus all we need to know is whether the number n is divisible by 9
Question: do the Digits of N sum to a Multiple of 9?
Or
Is: a + b + c + d + 15 = Multiple of 9?
S1: (a + b + c + d) = 9k + 3
Where k = non negative integer
Inserting this result into the question stem:
(9k + 3) + 15 = 9k + 18
Rule: when you add 2 multiples of some Number X —— the result will be another multiple of X
9k + 18 = Multiple of 9
Which means the digits of n must sum to a Multiple of 9, meaning n will be divisible by 9 and thus 36. YES
S1 is sufficient alone
S2: the digits (a+ b + c + d) sum to a multiple of 3
Case 1: they could sum to 6, in which case the sum of the digits of N = 21 and N would NOT be divisible by 9
Case 2: the digits could sum to 12, a multiple of 3, in which case the sum of the digits of N = 27 and N would be divisible by 9
Since we get conflicting answers of yes and no, s2 is not sufficient.
A: S1 alone is sufficient
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36 = (2)^2 * (3)^2
Divisibility Rule for (2)^n power
If the last n digits are divisible by (2)^n
Then the number will be divisible by (2)^n
As given, the last 2 digits of n are 96, which is divisible by 4
Thus all we need to know is whether the number n is divisible by 9
Question: do the Digits of N sum to a Multiple of 9?
Or
Is: a + b + c + d + 15 = Multiple of 9?
S1: (a + b + c + d) = 9k + 3
Where k = non negative integer
Inserting this result into the question stem:
(9k + 3) + 15 = 9k + 18
Rule: when you add 2 multiples of some Number X —— the result will be another multiple of X
9k + 18 = Multiple of 9
Which means the digits of n must sum to a Multiple of 9, meaning n will be divisible by 9 and thus 36. YES
S1 is sufficient alone
S2: the digits (a+ b + c + d) sum to a multiple of 3
Case 1: they could sum to 6, in which case the sum of the digits of N = 21 and N would NOT be divisible by 9
Case 2: the digits could sum to 12, a multiple of 3, in which case the sum of the digits of N = 27 and N would be divisible by 9
Since we get conflicting answers of yes and no, s2 is not sufficient.
A: S1 alone is sufficient
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18 Jan 2021, 10:13
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If n is a 6-digit positive integer, abcd96, where a is the hundred-thousands digit of n, b is the ten-thousands digit of n, c is the thousands digit of n, and d is the hundreds digit of n, is n divisible by 36?
Divisibility of 36 = 9*4= It must be divided by 9 and 4.
Stat1: The remainder is 3 when a+b+c+d is divided by 9.
So, sum of digits = 3+9+6 = 18, it will be divisibly by 9 and n had last 2 digit 96 is divisible by 4. So Sufficient
Stat2: a+b+c+d is divisible by 3.
Sum of a+b+c+d = 3, 6, 9....So sum of digits of n=18, 21, 24.... So divisible at 18 and not divisible at 21 or 24...Not sufficient.
So, I think A.
Divisibility of 36 = 9*4= It must be divided by 9 and 4.
Stat1: The remainder is 3 when a+b+c+d is divided by 9.
So, sum of digits = 3+9+6 = 18, it will be divisibly by 9 and n had last 2 digit 96 is divisible by 4. So Sufficient
Stat2: a+b+c+d is divisible by 3.
Sum of a+b+c+d = 3, 6, 9....So sum of digits of n=18, 21, 24.... So divisible at 18 and not divisible at 21 or 24...Not sufficient.
So, I think A.
