What is the tens digit of 7^1415?

Hi,
Another approach to this Q, which would be correct for all numbers..

the Q can be answered if we can find the remainder when \(7^{1415}\) is divided by 100..
now 7^4=2401..

\((7^4)^{353} * 7^3 = (2401)^{353}*7^3\)..
2401 will always leave a remainder of 1
so finally remainder = \(1* 7^3 = 1* 343\)..
or remainder = 43..
tens digit =4
E
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7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
We should see this as pattern recognition . We have a cycle of 4 . (We can multiply the last 2 digits only as we care about ten's digit )
0 , 4 , 4 , 0 .
1415= 4*353 + 3
The ten's digit will be 4 .
Answer E
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I am bit confused here

Clearly the cyclicity is 4 here.

7^1415

= 7 ^[4.353 + 3]
= (7 ^4.353) . (7 ^3)
= (7^4.(353)) . (7^3)
= (2401^353) . ( 7^3 )
= (Tens digit of a number that ends in 1 is equal to the last digit of power multiplied by the tens digit of the number) .(7^3)
= (0) . (343)
= 0
= The tens digit is 0.

Yes, the cyclicity is 4 but considering that we start from the exponent of 1, the cyclicity is
0, 4, 4, 0
0, 4, 4, 0
...
etc

So 1415 = 4*353 + 3 means 353 full cycles and 3 more so you go to 0, 4, 4
The digit will be 4.
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\(7^{1415}\) = \(7^{1410}*7\) = \(49^{705}*7\) = \(49^{704}*49\) = \(01^{352}*49\) = \(1*49=49\)
Tens digit is 4

Hi chetan2u, sorry, but why you are calculating (7^4)^{354} * 7^3? Shouldn't it be (7^4)^{353} * 7^3? Or am I missing something??

Hi VeritasKarishma gmatbusters Gladiator59 Bunuel EMPOWERgmatRichC

I adore chetan2u 's approach above. Is there a way to skip two digit multiplications since
they are too prone to error?
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Why to complicate with the various way to approach the same.

Lets understand the rule of cyclicity!

7 has a cyclicity of 4...i.e every 4th number the same pattern repeats

Now divide the exponent 1415 by 4, the balance would be 3.
now its just 7 the cube...i.e 7x7x7.....the unit digit can be easily found.

cyclicity rule is as below:

1 1
2 4
3 4
4 2
5 1
6 1
7 4
8 4
9 2
10 1

Apply the above yardstick and solve it accordingly.

When we divide a number by 100, the last 2 digits of the Dividend will be the Remainder.


Well, let’s try it out with the Chinese Remainder Theorem..... (not applicable on GMAT)


(7)^1415 /100 ———> Rem of = ?

Split up the denominator into: 50 * 2 = 100

(7)^1415 /50

(7^2)^707 * (7) /50

(49)^707 * (7) /50

(50 - 1)^707 * (7) /50 ——> Rem of = ?

Using the Concept of the Binomial Theorem, when we multiply out (50 - 1)^707 ——-> every term will be divisible by 50 except the last term of: (-1)^707 = -1

Excess remainder therefore =

(-1) * 7 = -7

-7 + (divisor of 50) = 43

n = 50a + 43

The given number has the characteristic that when it is divided by 50 ——-> yields a remainder of +43

AND

(7)^1415 /2 ——-> since the result will be Odd, it will yield a remainder of 1 when divided by 2

N = 2b + 1

And

N = 50a + 43

So, we are looking for the first ODD number that, when divided by 50, will yield a remainder of 43

That number is 43

Thus:


(7)^1415 when divided by 100 will yield a remainder of ——-> 43

The last 2 digits must be 43

And the tens digit must be 4

E

Posted from my mobile device

Can we use this same approach for :
a) any exponent
b) questions that ask for digits larger than tens ( e.g. hundreds )

Posted from my mobile device
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