If 3^10–n is divisible by 4, which of the following could be the value

chetan2u could you tell what is wrong with my reasoning.

3 power 10 would end in 0 - 5 would give another odd number. Hence only divisive by n=1

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3*10 would end in 0
But 3^10=3*3*3*3*...10times
Cyclicity
3^1 ends in 3
3^2 =9 ends in 9
3^3 = 27 ends in 7
3^4=27*3 ends in 1
3^5=81*3 ends in 3 and the cyclicity happens after every 4th number
So 3^10=3^(4*2+2) so will have same units digit as 3^2 so 9..
So 3^anything positive will always be ODD

Argh..what was I thinking??
Thanks mate.

Given \(3^{10}–n\) = \(4k\)

I. \(n = 0\), we get \(3^{10}–n\) = \(3^{10}\) = \(3^{4}*3^{4}*3^{2}\)

When \(3^{4}*3^{4}*3^{2}\) is divided by \(4\), we get remainder as \(1\). Hence Not Divisible.

\(n = 0\) is not possible.


II. \(n = 1\), we get \(3^{10}–n\) = \(3^{10}–1\)

we know from above \(3^{10}\) divided by \(4\) leaves a remainder of \(1\)

& \(1\) divided by \(4\) will leave a remainder of \(1\).

Hence \(3^{10}–1\) divided by 4 will leave a remainder \((1 - 1) = 0\). Hence divisible.

\(n = 1\) is possible.


III. \(n = 5\), we get \(3^{10}–n\) = \(3^{10}–5\)

we know from above \(3^{10}\) divided by \(4\) leaves a remainder of \(1\)

& \(5\) divided by \(4\) will leave a remainder of \(1\).

Hence \(3^{10}–5\) divided by 4 will leave a remainder \((1 - 1) = 0\). Hence divisible.

\(n = 5\) is possible.


Answer D.


Thanks,
GyM

If you don't subtract anything than 3^10-0 will be odd hence option 1 and 5 can be eliminated immediately.

Now 3^10 is a big number, say x. If either 1 or 5 is subtracted from a big number the both the numbers will be divisible by 4 if either one of them is divisible by 4. (e.g. 9-1 = 8 & 9-5=4; BOTH 8 & 4 are divisible by 4)

Since there is no option which says none of these then we can safely select option D with doing any calculations!! Hope this out of the box thinking was helpful & time saving

3^10 - n/ 4 = Q

So 3's cycle of power is 3,9,7,1...if we raise 3 to 10 times we will get a number that ends in XX9.
So 9 - 5 = 4 = divisible
9 - 8 = divisible.

So answer is II,III which is D

3=3
3*3 =9
3*3*3=27
3*3*3*3=81

then the last digit repeats itself.
so starting from 3 count 10 times we come to 9(last digit)

then we know that the last digit for 3^10 is 9.
and this -n is divisible by 4.that means n will be 1 or 5.

can we reason as follows:

since 3^10 is odd, 3^10 -1 is even (odd-odd gives even). and since we get an even result, it could be divisible either by 4 or 5 ?

thanks

3^10=729^2= 29*29 = a number ending with 41. Therefore:
1) 41-0 = 41 not divisible by 4
2) 41-1 = 40 divisible by 4
3) 41-5 = 36 divisible by 4

3^10 = (4-1)^10 = 4k+1, where k is an integer.
So any n = 4m+1 will work, where m is an integer. Only 1 and 5 can be represented as 4m+1.

Posted from my mobile device

\(\frac{3^{10} – n}{4}\) = 0

Or, \(\frac{3^2*3^2*3^2*3^2*3^2 – n}{4}\) = 0

Or, \(\frac{9}{4}*\frac{9}{4}*\frac{9}{4}*\frac{9}{4}*\frac{9}{4}-\frac{n}{4}\) = 0

Or, \(\frac{81}{4}*\frac{729}{4}-\frac{n}{4}\) = 0

Or, \(\frac{59049}{4}-\frac{n}{4}\) = 0

Or, \(\frac{59049 - n }{4}\) = 0

Now, chck the options only (II) and (III) fits in perfectly, Answer must hence be (D)

chetan2u
i could not fully comprehend the point (ii) of your explanation. Can you please illustrate using a example and can it be used with other expressions also for eg 5^11/4?

The exponents of 3 follow this pattern in the units digit = 3, 9, 7, 1

3^10 = units digit of 9.

Lets take a look at the options:

I. 0; 9 - 0 is not divisible by 4. OUT.
II. 1; 9 - 1 is divisible by 4. In.
III. 5; 5 - 1 is divisible by 4. In.

Answer is D.

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