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14 May 2021, 01:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (02:19) correct
31%
(01:52)
wrong
based on 84
sessions
History
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Is the tens digit of a three-digit positive integer p divisible by 3?
(1) p - 7 is a multiple of 3
(2) p - 13 is a multiple of 3
(1) p - 7 is a multiple of 3
(2) p - 13 is a multiple of 3
17 May 2021, 01:15
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Bunuel
Is the tens digit of a three-digit positive integer p divisible by 3?
(1) p-7 is a multiple of 3 --> \(p=3x+7=3(x+2)+1\). p is 1 more than a multiple of 3. Not sufficient.
(2) p-13 is a multiple of 3 --> \(p=3y+13=3(y+4)+1\). p is 1 more than a multiple of 3. Not sufficient.
(1)+(2) Nothing new. For example, consider p=103 for an YES answer or p=112 for a NO answer. Not sufficient.
Answer: E.
Hope it's clear.
General Discussion
WannaManaConsult
14 May 2021, 15:07
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I approached this by using the rule that the sum of the digits must be equal to a multiple of 3 for a number to be a multiple of 3.
so let p=abc, where abc are digits.
The question then is: is b divisible by 3?
1)
In the cases of c=7, 8, or 9, b remains unchanged and we basically have
a+b+c-7 = a multiple of 3.
In cases where c=0 to 6, the digits become (a, b-1, 10+c-7) and we have
a+(b-1)+(10+c-7)=a+b+c+2 = a multiple of 3
This is insufficient to determine if b is 0, 3, 6, or 9 because
(a,b,c) can be, for example, (3,6,7) and (4,5,7) and satisfy the condition p-7 is a multiple of 3.
Insufficient.
2)
Analysis is similar for p-13=multiple of 3
In cases where c>=3, and b>=1, the digits become (a,b-1,c-3)
c>=3 and b=0, (a-1,9,c-3)
c<3 and b>=2, (a,b-2,10+c-3)
c<3 and b<=2, (a-1,10+b-2,10+c-3)
Notice we know nothing about a other than 1 is subtracted from it in the case of c>=3 and b=0. The problem is for all other cases, a can be adjusted up or down to make b a multiple of 3, or not.
Both together is still insufficient for the same reasons.
so let p=abc, where abc are digits.
The question then is: is b divisible by 3?
1)
In the cases of c=7, 8, or 9, b remains unchanged and we basically have
a+b+c-7 = a multiple of 3.
In cases where c=0 to 6, the digits become (a, b-1, 10+c-7) and we have
a+(b-1)+(10+c-7)=a+b+c+2 = a multiple of 3
This is insufficient to determine if b is 0, 3, 6, or 9 because
(a,b,c) can be, for example, (3,6,7) and (4,5,7) and satisfy the condition p-7 is a multiple of 3.
Insufficient.
2)
Analysis is similar for p-13=multiple of 3
In cases where c>=3, and b>=1, the digits become (a,b-1,c-3)
c>=3 and b=0, (a-1,9,c-3)
c<3 and b>=2, (a,b-2,10+c-3)
c<3 and b<=2, (a-1,10+b-2,10+c-3)
Notice we know nothing about a other than 1 is subtracted from it in the case of c>=3 and b=0. The problem is for all other cases, a can be adjusted up or down to make b a multiple of 3, or not.
Both together is still insufficient for the same reasons.
