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If x is a positive integer, is the remainder 0 when [#permalink]

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15 Aug 2009, 18:32

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Please, show your logic. Thanks, -----------------------------

Q32: If x is a positive integer, is the remainder 0 when (3^x + 1)/10? (1) x = 3n + 2, where n is a positive integer. (2) x > 4

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.

Last edited by TriColor on 16 Aug 2009, 09:36, edited 1 time in total.

If x is a positive integer, is the remainder 0 when (3x + 1)/10? - divisibility by 10 means 3x+1 has zero as a last digit.

(1) x = 3n + 2, where n is a positive integer. Let's just substitute x for this expression: 3*(3n+2) + 1 = 9n + 7. In order to have 0 as a last digit. 9n must end with digit 3, otherwise our expression cannot be divisible by 10. It's possible for n=7: 9*7=63. Therefore, the expression can be divisible by 10 (n=7) and cannot be divisible by 10 (n=6). Insufficient.

(2) x > 4 Insufficient along and doesn't add any support to first statement.
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Please, show your logic. Thanks, -----------------------------

Q32: If x is a positive integer, is the remainder 0 when (3x + 1)/10? (1) x = 3n + 2, where n is a positive integer. (2) x > 4

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.

Agree with E. 1) if x=3n+2, then 3x+1 = 9n+7. If n=7 , then 9n+7/10 will have 0 as remainder. For other values remainder will be other than 0. Not suff. 2) x>4. Now, we need to find some multiple of 3 to which when 1 is added the result is divisible by 10. x can be 33 in which case remainder is 0 , for some other values remainder will not be 0. Not suff.

Combining, we have 9n+7 / 10 where n>2/3. So n can take any value which might satisfy the equation. E.

Hello, I have correcter the question... X should be the exponent of 3... Sorry about the mistake. -----------------------------

Q32: If x is a positive integer, is the remainder 0 when (3^x + 1)/10? (1) x = 3n + 2, where n is a positive integer. (2) x > 4

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.

EDITED: Sometimes I also make silly mistakes... I corrected my post.

it is old DS-trap here. After good preparation you can solve this in 10-15 sec. Here is a pattern:

1. The question about divisibility/last digit 2. The question contains exponent a^x 3. x can be periodical (x=a*x+b, one of the statement). In other words, something like: each 4th number and so on. 4. other statement is obviously insufficient.

Where is a key? Just write out last digit for a few consecutive numbers of exponent:

3^x x=0: 1 x=1: 3 x=2: 9 x=3: 7 x=4: 1 period:4

Let's consider any other example:

7^x x=0: 1 x=1: 7 x=2: 9 x=3: 3 period:4

6^x x=0: 1 x=1: 6 x=2: 6 x=3: 6 period:1 (need to check for x=0)

2^x x=0: 1 x=1: 2 x=2: 4 x=3: 8 x=4: 6 x=3: 2 period:4 (need to check for x=0)

So, for any a^x exponent there is a pattern and GMAC is trying to trick you here. But if you can instantly recognize the DS-pattern, you in a few second will "feel" that answer is A and will spend next 10 seconds to check match between 3^x period for last digit (here it is 3) and period for one statement, (3n+2) has period 4. They don't equal. So the answer is E. You even may not know the answer (divisible or not) you only know for sure that last digit of 3^x+1 is always the same.
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great problem. guys, can anyone write down a general rule on how to determine a last digit when a number raised to power? Thanks! I took the right approach but could not avoid using a calculator

great problem. guys, can anyone write down a general rule on how to determine a last digit when a number raised to power? Thanks! I took the right approach but could not avoid using a calculator

Its just same as Walker mentioned. You gotta remember the pattern for 3 , 4, 6, 7, 9 and 11 powers of x. I have not seen any others being tested

great problem. guys, can anyone write down a general rule on how to determine a last digit when a number raised to power? Thanks! I took the right approach but could not avoid using a calculator

You don't need to do all math here. Let's say, you want to see last digit of 353^x. The main rule here:

last digit of 353^x is equal to last digit of 3^x. So, you can always cut all digits but last.