S97-15

Question

The tens digit of  6^{17}  is

A. 1
B. 3
C. 5
D. 7
E. 9

Bunuel · Answer

Official Solution:

The tens digit of  6^{17}  is

A. 1
B. 3
C. 5
D. 7
E. 9

We know that there must be a pattern, since we can’t be expected to expand  6^{17}  out to all its digits. In other words, we must be able to spot a repeating cycle of digits.
 
The only way forward is to compute tens digits for powers of 6, starting with  6^1 , and see what we get. To go up, multiply the previous result by 6 and drop any higher digits than the tens, but we have to keep the units digit (which, as we’ll see, will be 6 every time).
 
 6^1 = 6  (no tens digit)
 
 6^2 = 6 	imes 6^1 = 36  (tens digit = 3)
 
 6^3 = 6 	imes 6^2 = ..16  (tens digit = 1)
 
 6^4 = 6 	imes 6^3 = ..96  (tens digit = 9)
 
 6^5 = 6 	imes 6^4 = ..76  (tens digit = 7)
 
 6^6 = 6 	imes 6^5 = ..56  (tens digit = 5)
 
 6^7 = 6 	imes 6^6 = ..36  (tens digit = 3)
 
Whew - the numbers finally started repeating! The cycle is 3, 1, 9, 7, 5 - which is 5 terms long. Every power will have the same tens digit as the 5th larger power, so  6^2 ,  6^7 ,  6^{12} , and most importantly  6^{17}  will all have 3 as their tens digit.
 
Notice that the pattern didn’t start until  6^2 .  6^1  doesn’t have a tens digit (or has a tens digit of 0, but this digit is never repeated later in the cycle).

Answer: B

gauravmishra132 · Answer

It can also be done fast by breaking 6 into: 2 to the power 17 and 3 to the power 17. and then check the cyclicity....

ng90 · Answer

Can you please explain how?

Bunuel · Answer

What is the tens digit of 6^17?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

There are several ways to deal with this problems some easier some harder, but almost all of them are based on the pattern recognition.

The tens digit of 6 in integer power starting from 2 (6^1 has no tens digit) repeats in pattern of 5: {3, 1, 9, 7, 5}:
The tens digit of 6^2=36 is 3;
The tens digit of 6^3=216 is 1;
The tens digit of 6^4=...96 is 9 (how to calculate: multiply 16 by 6 to get ...96 as the last two digits);
The tens digit of 6^5=...76 is 7 (how to calculate: multiply 96 by 6 to get ...76 as the last two digit);
The tens digit of 6^6=...56 is 5 (how to calculate: multiply 76 by 6 to get ...56 as the last two digits);
The tens digit of 6^7=...36 is 3  again  (how to calculate: multiply 56 by 6 to get ...36 as the last two digits).

Hence, 6^2, 6^7, 6^12, 6^17, 6^22, ... will have the same tens digit of 3.

Answer: B.

shivamlohiya007 · Answer

any alternate solution

Bunuel · Answer

Have you checked post just above yours? https://gmatclub.com/forum/s97-184703.html#p1974800

rohityagami25 · Answer

This can also be done using the concept of "finding the last two digits" of number system.
6*6^16 = 6 * ((6^2)^8) = 6 * (36^8)

R1) Now in finding the last two digits there is a rule that states the last two digits of (50 + k)^2 is same as last two digits of k^2.
So last two digits of 36^8 is same as (14^2)^4 = 96^4

R2) Now another rule says, last two digits of (100-k)^2 is same as k^2.
So last two digits of 96^4 is same as (4^2)^2 = 16 ^2 = 256

Therefore 6*6^16 = 6 * 56 = 36
So ten's digit is 3.

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