If p, q and r are digits, is (p + q + r) a multiple of 9? (1) The thr

Given: p, q and r are digits.

Asked: Is (p + q + r) a multiple of 9?

(1) The three digit number pqr is a multiple of 9
100p + 10q + r is multiple of 9
(p+q+r) is a multiple of 9
SUFFICIENT

(2) (p x q) + r is a multiple of 9
p*q + r is a multiple of 9
Case 1: p=2 q=3 r=3 => pq+r = 2*3+3= 9 is a multiple of 9 but p+q+r= 2+3+3=8 is not a multiple of 9
Case 2: p=9 q=0 r=9 => pq+r = 9*0+3= 9 is a multiple of 9 and p+q+r= 9+0+9=18 is a multiple of 9
NOT SUFFICIENT

IMO A

Hi everybody,

If p, q and r are digits, is (p + q + r) a multiple of 9?


(1) The three digit number pqr is a multiple of 9
in order for par to be a multiple of 9 the same of its digits must be divisible by 9.

Hence statement 1 is sufficient

(2) (p x q) + r is a multiple of 9
In this case we have different outcomes

Consider p=2,q=3 and r=3. statement 2 is satisfied but the sum of the three digit is not a multiple of 9
consider p=9,q=9 and r=9. statement 2 is satisfied and the sum of these digits is a multiple of 9

Hence statement 2 is not sufficient

Option A is the correct answer

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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Condition 1)
Remind that if a number is a multiple of \(9\), then the sum of all digits is a multiple of \(9\).
The condition that the three digit number \(pqr\) is a multiple of \(9\) is equivalent to the question that \(p + q + r\) is a multiple of \(9\).
Thus condition 1) is sufficient.

Condition 2)
If \(p = 9\), \(q = 9\) and \(r = 9\), then \(p + q + r = 27\) is a multiple of \(9\) and the answer is 'yes'.
If \(p = 1\), \(q = 2\) and \(r = 7\), then \(p + q + r = 10\) is not a multiple of \(9\) and the answer is 'no'.
Since condition 2) does not yield a unique answer, it is not sufficient.

Therefore, A is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.