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19 Jun 2023, 01:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (02:19) correct
38%
(01:57)
wrong
based on 42
sessions
History
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Not Attempted Yet
If M and N are positive integers greater than 1, what is the remainder when the expression \(22^{3M} * 39^{2N} + 14^{2(M+N)}\) is divided by 5?
(1) M = 13
(2) N = 14
(1) M = 13
(2) N = 14
19 Jun 2023, 02:05
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The question asks us to find the remainder when the expression 22^(3M) * 39^(2N) + 14^(2(M+N)) is divided by 5, given that M and N are positive integers greater than 1.
To find the remainder when the expression is divided by 5, we only need to look at the last digit of each term. The last digit of 22^k repeats in the pattern 2, 4, 8, 6, 2, 4, 8, 6, .... The last digit of 39^k repeats in the pattern 9, 1, 9, 1, .... The last digit of 14^k repeats in the pattern 4, 6, 4, 6, ....
Since 3M is always a multiple of 3, the exponent of 22 is always a multiple of 3. Therefore, the last digit of 22^(3M) is always 8.
Since 2N is always even, the exponent of 39 is always even. Therefore, the last digit of 39^(2N) is always 1.
Since 2(M+N) is always even, the exponent of 14 is always even. Therefore, the last digit of 14^(2(M+N)) is always 6.
Adding the last digits of each term, we get 8 + 1 + 6 = 15. Therefore, the remainder when the expression is divided by 5 is 0, regardless of the value of M.
Since both statements give a specific value for M, we can calculate the remainder when the expression is divided by 5. Therefore, both statements are sufficient to answer the question.
To find the remainder when the expression is divided by 5, we only need to look at the last digit of each term. The last digit of 22^k repeats in the pattern 2, 4, 8, 6, 2, 4, 8, 6, .... The last digit of 39^k repeats in the pattern 9, 1, 9, 1, .... The last digit of 14^k repeats in the pattern 4, 6, 4, 6, ....
Since 3M is always a multiple of 3, the exponent of 22 is always a multiple of 3. Therefore, the last digit of 22^(3M) is always 8.
Since 2N is always even, the exponent of 39 is always even. Therefore, the last digit of 39^(2N) is always 1.
Since 2(M+N) is always even, the exponent of 14 is always even. Therefore, the last digit of 14^(2(M+N)) is always 6.
Adding the last digits of each term, we get 8 + 1 + 6 = 15. Therefore, the remainder when the expression is divided by 5 is 0, regardless of the value of M.
Since both statements give a specific value for M, we can calculate the remainder when the expression is divided by 5. Therefore, both statements are sufficient to answer the question.
19 Jun 2023, 07:05
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Pre-think
The unit digit for 39^(2N) is always 1
The unit digit for 14^(2(M+N)) is always 6
(1) Only
M=13
3M=39
So we can get the unit digit for 22^(3M) and thus answer the remainder question.
Sufficient
(2) Only
If M=1, the unit digit for 22^3 is 8, and the answer to the remainder question is 4.
If M=2, the unit digit for 22^6 is 4, and the answer to the remainder question is 0.
Not sufficient.
The answer is thus (A).
The unit digit for 39^(2N) is always 1
The unit digit for 14^(2(M+N)) is always 6
(1) Only
M=13
3M=39
So we can get the unit digit for 22^(3M) and thus answer the remainder question.
Sufficient
(2) Only
If M=1, the unit digit for 22^3 is 8, and the answer to the remainder question is 4.
If M=2, the unit digit for 22^6 is 4, and the answer to the remainder question is 0.
Not sufficient.
The answer is thus (A).
