If x and y are positive integers, what is the remainder when

Question

If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?

(1) x = 25.
(2) y = 1.

Bunuel · Accepted Answer

Theory: First of all note that the last digit of xyz^(positive integer) is the same as that of z^(positive integer); • Integers ending with 0, 1, 5 or 6, in the positive integer power, have the same last digit as the base (cyclicity of 1): xyz0^(positive integer) ends with 0; xyz1^(positive integer) ends with 1; xyz5^(positive integer) ends with 5; xyz6^(positive integer) ends with 6; • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4; For example last digit of xyz3^{positive \ integer} repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. • Integers ending with 4 and 9 have a cyclicity of 2. Now, last digit of xyz3^{positive \ integer} repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... means that: 3^1, 3^5, 3^9, 3^13, ..., 3^(4x+1), all will have the same last digit as 3^1 so 1; 3^2, 3^6, 3^10, 3^14, ..., 3^(4x+2), all will have the same last digit as 3^2 so 9; 3^3, 3^7, 3^11, 3^15, ..., 3^(4x+3), all will have the same last digit as 3^3 so 7; 3^4, 3^8, 3^12, 3^16, ..., 3^(4x), all will have the same last digit as 3^4 so 1; So to get the last digit of 3^x, (where x is a positive integer) you should divide x by 4 (cylcility) and look at the remainder: If remainder is 1 then the last digit will be the same as for 3^1 (so the first digit from the pattern {3, 9, 7, 1}); If remainder is 2 then the last digit will be the same as for 3^2 (so the second digit from the pattern {3, 9, 7, 1}); If remainder is 3 then the last digit will be the same as for 3^3 (so the third digit from the pattern {3, 9, 7, 1}); If remainder is 0 then the last digit will be the same as for 3^4 (so the fourth digit from the pattern {3, 9, 7, 1}); Next, as 4+4x (the power of 3^(4+4x)) is clearly divisible by 4 (remainder 0) then the last digit of 3^(4+4x) is the same as the last digit of 3^4 so 1. You can apply this to integers ending with other digits as well (with necessary modification of pattern). Hope it's clear. P.S. Check this for more: math-number-theory-88376.html

Bunuel · Answer

If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? Last digit of 3^{positive \ integer} repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. Now, 3^{4+4x}=3^{4(1+x)} will have the same last digit as 3^4 which is 1 (remainder upon division the power 4+4x by cyclicity 4 is 0, which means that 3^{(4+4x)} will have the same last digit as 3^4). Now 3^{(4 + 4x)}+9^y will be divisible by 10 if y=odd , in this case the last digit of 9^{odd} will be 9 (9 has a cyclicity of 2: {9, 1} - {9, 1} - ...) so the last digit of 3^{(4 + 4x)}+9^y will be 1+9=0. (1) x = 25. Not sufficient. (2) y = 1 --> y=odd -> remainder=0. Sufficient. Answer: B. Similar question: divisiblity-with-109075.html Check Number Theory chapter of Math Book for more: math-number-theory-88376.html

subhashghosh · Answer

HI Bunuel

What does this mean ?

remainder upon division the power 4+4x by cyclicity 4 is 0, which means that 3^{(4+4x)} will have the same last digit as 3^4)

Regards,
Subhash

ankur1901 · Answer

Its a value question and hence we need a definate value.

3^(4+4x) will be 3^8, 3^12, 3^16....when x = 1, 2, 3.....
So we can safely conclude that 3^(4 + 4x) will always have remainder of 1(based on cyclicity) . So value of X does not matter.

Hence 1 is not sufficient - or does not help us to respond uniquely.

Second statement tells us that Y=1, which is exactly what need to reply. We dont have to go into the details of calculations to find out the remainder.
Answer would be B

CelestialBro · Answer

Here's the trick: "remainder when divided by 10" is just a sneaky way of asking for the last digit of the number.
 1. The 3^(4+4x) part:  Powers of 3 repeat their last digits in a cycle of 4 (ends in 3, 9, 7, then 1). Because the exponent (4+4x) is always a multiple of 4, it will  always  land on that 4th step. So, 3^(4+4x) will always end in a 1. The value of x doesn't matter at all!
 2. The 9^y part:  Powers of 9 just bounce between ending in 9 (odd power) and 1 (even power). To solve the problem, we literally only need to know if y is odd or even.
 
  Statement 1:  Tells us x = 25. Useless, since we already know the first part ends in 1 regardless of x. 
  Statement 2:  Tells us y = 1 (odd). This means 9^y ends in 9.  
Add their last digits together: 1 + 9 = 10. The final number ends in a 0, so the remainder is 0. Statement 2 alone solves it. (Answer is B)

If x and y are positive integers, what is the remainder when

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If x and y are positive integers, what is the remainder when

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