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07 May 2026, 20:16
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Difficulty:
55%
(hard)
Question Stats:
15% (01:13) correct
85%
(01:27)
wrong
based on 13
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y^4 is a two-digit positive odd integer. A positive integer n is divisible by y^2. What could be the sum of digits of n?
I. 1
II. 9
III. 17
A. I and II only
B. II only
C. II and III only
D. I and III only
E. I, II, and III
I. 1
II. 9
III. 17
A. I and II only
B. II only
C. II and III only
D. I and III only
E. I, II, and III
07 May 2026, 20:35
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Since y^4 is a two-digit positive odd integer, let’s find possible values of y.
Check odd integers:
1^4 = 1 → not two-digit
3^4 = 81 → two-digit and odd
5^4 = 625 → too large
So the only possible value is y = 3.
Then:
y^2 = 3^2 = 9
We are told n is divisible by y^2, so n must be divisible by 9.
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
Now test the choices:
I. 1 → not divisible by 9
II. 9 → divisible by 9
III. 17 → not divisible by 9
Only II could be the sum of digits of n.
Answer: B. II only
Check odd integers:
1^4 = 1 → not two-digit
3^4 = 81 → two-digit and odd
5^4 = 625 → too large
So the only possible value is y = 3.
Then:
y^2 = 3^2 = 9
We are told n is divisible by y^2, so n must be divisible by 9.
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
Now test the choices:
I. 1 → not divisible by 9
II. 9 → divisible by 9
III. 17 → not divisible by 9
Only II could be the sum of digits of n.
Answer: B. II only
Do we know that y is an integer ? We know that y^4 is one, and from the context y^2 should be, though this is not stated. y^2 could be 5, 7, or 9.
If y^2 is 5, n could be 10
If y^2 is 7, n could be 63 or 98
So the sum of the digits of n could any be 1,9, or 17 (among other possibilities)
If y^2 is 5, n could be 10
If y^2 is 7, n could be 63 or 98
So the sum of the digits of n could any be 1,9, or 17 (among other possibilities)
