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Re: If y is divisible by the square of an even prime number and x is the a [#permalink]

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14 Jan 2015, 15:38

Since y/4 is an integer (no remainder) then y is a multiple of 4. y=4,8,12,16 etc etc x=2^2=4

So, we are looking for sth like 4^4, 0r 4^8. We could already see that the units digit is6,(4^4=16), but let's test it using the cyclicity of 4: 4^1=4 4^2=16 4^3=64 here the patterns repeats itself.

So, every even power results in a units digit of 6, ANS D.

Re: If y is divisible by the square of an even prime number and x is the a [#permalink]

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29 May 2017, 10:39

Bunuel wrote:

Tough and Tricky questions: Properties of Numbers.

If y is divisible by the square of an even prime number and x is the actual square of an even prime number, then what is the units digit of x^y?

A. 0 B. 2 C. 4 D. 6 E. 8

This problem is best tackled via making a table.

We know that x = 4 since it's the square of the only even prime, 2. We know that y is a multiple of 4.

Y | 4^y 0 | 1 --> technically 0 is a multiple of any number, but since 1 is not in the answer choices we can ignore. 4 | = 2^8 = 256 8 | 2^16 12 16

At this point, the numbers become too big to compute in a reasonable amount of time, but since it's a PS problem we know only one value can be correct, so as long as we find this one value (4^4 = 256), then we are all set and do not need to compute any other values in the table. The units digit is 6.

Re: If y is divisible by the square of an even prime number and x is the a [#permalink]

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16 Nov 2017, 08:41

I don't understand why the exponent(y) musst be 4. Because every multiple of 4 is devisible by 4. So y could be 4,8,16...? So why do we have to take just even squares of 4? "If y is divisible by the square of an even prime number...." Yes the number is 2, the square is 4. But why can y not be every multiple of 4? Than the answer could be C or D.. Can someone explain please?

If y is divisible by the square of an even prime number and x is the a [#permalink]

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16 Nov 2017, 09:33

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Bunuel wrote:

Tough and Tricky questions: Properties of Numbers.

If y is divisible by the square of an even prime number and x is the actual square of an even prime number, then what is the units digit of x^y?

A. 0 B. 2 C. 4 D. 6 E. 8

Dokami wrote:

I don't understand why the exponent(y) musst be 4. Because every multiple of 4 is devisible by 4. So y could be 4,8,16...? So why do we have to take just even squares of 4? "If y is divisible by the square of an even prime number...." Yes the number is 2, the square is 4. But why can y not be every multiple of 4? Than the answer could be C or D.. Can someone explain please?

Dokami ,I think you might be confusing what \(y\) can have as its units digit with what \(x^{y}\) can have as its units digit.

x is the actual square of an even prime number. There is only one even prime: 2. x = 4

y is divisible by the square of an even prime number: y is divisible by 4; y is a multiple of 4.

y can be any multiple of 4.

We need the units digit of \(x^{y}\). That means \(4^{y}\). We are dealing with the cycles created by 4 raised to powers ("cyclicity"). Cyclicity determines units digits.

Look at the units digits. They alternate. Odd powers of four end in 4. Even powers of four end in 6. In fact, when listing, in order, numbers to powers, you need only to multiply the units digits.

If the exponent is even, 4 raised to that even power ends in 6. Every multiple of 4 is even. \(x^{y}\), where \(x\) = 4 and \(y\) is a multiple, will have a units digit of 6.

Re: If y is divisible by the square of an even prime number and x is the a [#permalink]

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17 Nov 2017, 06:12

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I think I got it. Please correct me if I am wrong: Because y could be 4,8,12... we have to check the solutions for 4^4, 4^8, 4^12,.... And because 4,8,12... are even, we have to check the solutions for cyclicity like you have done above. And here we can see that the units digit is "6", if we rise a even number.

If y is divisible by the square of an even prime number and x is the a [#permalink]

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17 Nov 2017, 08:29

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Dokami wrote:

I think I got it. Please correct me if I am wrong: Because y could be 4,8,12... we have to check the solutions for 4^4, 4^8, 4^12,.... And because 4,8,12... are even, we have to check the solutions for cyclicity like you have done above. And here we can see that the units digit is "6", if we rise a even number.

Very, very close! Just one little change: You wrote, "we have to check the solutions for 4^4, 4^8, 4^12."

You do not have to check more than one multiple of 4, where 4 = y = exponent. What matters is the pattern created by x = 4 raised to a power: Odd exponent? \(x^{y}\) ends in 4 Even exponent?\(x^{y}\) ends in 6. Every single multiple of 4 is even. If we raise 4 (x = 4) to any even power, the units digit is 6.

I might misunderstand you. Maybe we are saying the same thing. (It is a bit confusing because we have two 4s: x, the base, and y, the multiple of 4.)

I just do not want you to think you have to check more than one multiple of 4 when 4= exponent = y = multiple of 4. You don't. All exponents that are multiples of four are even, so they all end in 6, and you could be there all day. The highlighted part of what you wrote is what matters. Nice work.