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18 Feb 2011, 12:00
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B
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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.
(1) x = 25.
(2) y = 1.
19 Feb 2011, 03:11
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subhashghosh
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
18 Feb 2011, 12:30
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banksy
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
