The last digit of 12^12 + 13^13 14^14×15^15 =

Question

The last digit of  12^{12} + 13^{13} – 14^{14}*15^{15} =

A. 0
B. 1
C. 5
D. 8
E. 9

Bunuel · Accepted Answer

VERITAS PREP OFFICIAL SOLUTION:

Solution: B

Units digits of exponents are cyclical, and once you’re determined the full cycle of a particular units digit you can simply extrapolate its pattern forward. Checking these cycles for the units digits of 2, 3, 4, and 5 above: we know that units digits of powers of 2 follow a cycle of 2, 4, 8, 6. Therefore the units digit of 12^12 is 6. The last digits of powers of 3 follow a cycle of 3, 9, 7, 1. Therefore the units digit of 13^13 is 3. The units digits of 4 follow a cycle of 4, 6. Hence the units digit of 14^14 is 6. The units digit of any power of 5 remains 5 so the last digit of 15^15 will be 5.

This implies that last digit of 12^12+13^13 will be 9 and that of 14^14×15^15 will be 0. Since 14^14×15^15 is definitely much greater than (12^12+13^13),(12^12+13^13)–(14^14×15^15) will be a negative number with a units digit of 1. Especially if you selected 9 as your answer, remember this little GMAT trick well – for units digit problems that involve subtraction, beware the case in which the second number will be larger than the first, in which case a negative number will result!

peachfuzz · Answer

Last digit of:

12^12 is 6
13^13 is 3
14^14 is 6
15^15 is 5

6 + 3 - 6 x 5
=-21

1 is the last digit

Answer: B

Lucky2783 · Answer

This question tests the cyclicity concept. 
2 has a cyclicity of 4 i.e. 5th power of 2 has same unit digit as 1st power of 2 .
3 also has cyclicity of 4
4 has cycling of 2
5 has 0.

So this complicated calculation boils down to

2^12 +3^13 - 4^14 ×5^15

6+3-6×0=9 as unit digit is answer.

Answer E

gamerssc · Answer

Using cyclicity of 2^12 will have unit digit as 6 (12/4=3 no remainder, 2^4 = 16, so 6)
similarly for 3^13 the unit digit is 3
For 4^14 unit digit is 6
For 5^15 unit digit is 5
Now ( 6 + 3 - (6*5)) = 9-( 30) => 9-0 (considering only unit digit) = 6

KarishmaB · Answer

There are 3 terms here:

12^{12} 
 + 13^{13} 
 – 14^{14}×15^{15}  - This term is very simple to manage in case of last digit. An even number multiplied by a multiple of 5 will end in 0. Also, note that it is a huge term as compared with the other two terms.

12^{12}  - 2 has a cyclicity of 2 - 4 - 8 - 6 so 12 2s will end in 6
 13^{13}  - 3 has a cyclicity of 3 - 9 - 7 - 1 so 13 3s will end in 3

(A number ending in 6) + (A number ending in 3) gives a number ending in 9.

So we have 
Number ending in 9 - Number ending in 0 (much bigger) so the result will end in 1 and will be negative.

Answer (B)

Devlikes · Answer

Hi,
In order to make this concept more clear, I have a query: 
If we had
Number ending in 9 - Number ending in 3 (much bigger), then the answer would have been 4 in negative and not 6. 
Am I correct ?

chetan2u · Answer

hi,
yes you are right..
and the number need not be much bigger but even just bigger than left side so as to give us a negative number is enough to give 4 as remainder..

MruDe · Answer

Just take a number greater than 9, that ends in 3. for ex. 13. So, 13-9=4.

like in 0..take 10, so 10-9=1

KarishmaB · Answer

How do you subtract one number out of another? The one with the greater absolute value goes on the top and the smaller absolute value goes under it. You subtract and the result gets the sign of the greater absolute value.

100 - 29 ------>

100
-29
------
071
------

29 - 100 ------->

100
-29
-------
071
------

But since the sign of 100 is negative, your answer is -71.
So the number with greater absolute value is always on top.

Number ending in 9 - Number ending in 0 (much bigger) will look like------->

.........0
- ......9
--------
..........1
---------
Answer will be negative ending in 1.

Number ending in 9 - Number ending in 3 (much bigger) will look like ------->

............3
- .........9
--------
............4
---------
Answer will be negative ending in 4.

MBAPursuit123 · Answer

What a savage trap. Totally fell for this.

Funsho84 · Answer

me too.

I got all of the units digits but instead of multiplying the last 2 digits, I added them. I got 11-11 = 0.

A silly mistake is the worst feeling.

shridhar786 · Answer

to find 12^12 last digit
since we are concerned with last digit we can take it 2^12=2^4*2^4*2^4
 last digits =6*6*6=6
 for 13^13 last digit=3^4*3^4*3^4*3
 =1*1*1*3=3
 - in 14^14×15^15 we will get 2 and 5 as a factor so last digit will be zero
 now 12^12 + 13^13 – 14^14×15^15
 6+3-0=9-0
 since 14^14×15^15 is bigger number than 12^12 + 13^13
 so here (..............0)-(.........9)=1
 bigger no smaller no

Abhishek009 · Answer

12^{4*3}  will have unit's digit 6

13^{4*3}  will have unit's digit 1 &  13^1  will have units digit 3, So,  13^{13}  will have units digit 3

14^{14}  will have unit's digit 6

15^{15}  will have unit's digit 5

So, The units digit of  12^{12} + 13^{13} – 14^{14}×15^{15}  will have units digit ,  6 + 3 - 6*5 = 9 - 30  =  xx1

Thus, the units digit will be (B) 1, Answer.

leoxcvi · Answer

Can someone explain the practical purpose of such a question on the GMAT? Thanks.

Kinshook · Answer

Asked: The last digit of  12^{12} + 13^{13} – 14^{14}*15^{15} =

Last digit of 12^12 = 6
Last digit of 13^13 = 3
Last digit of 14^14 = 6
Last digit of 15^15 = 5

The last digit of  12^{12} + 13^{13} – 14^{14}*15^{15} = 6 + 3 - 6*5 = 9 - 30 = -21 
Since 14^{14}*15^{15} is larger than 12^{12} + 13^{13} , the result will be a negative number.

IMO B

samarpan.g28 · Answer

The cyclicity of 2 is 4 {2,4,8,6}. 12 is divisible by 4 therefore the unit digit of  12^{12}  is 6.
The cyclicity of 3 is 4 {3,9,7,1}. 13 is not divisible by 4 as it leaves remainder=1, therefore the unit digit of  13^{13}  is 3.
The cyclicity of 4 is 2 {4,6}. 14 is divisible by 2 therefore the unit digit of  14^{14}  is 6.
The cyclicity of 5 is 1 {5}. 15 is divisible by 1 therefore the unit digit of  15^{15}  is 1.
​
Now, 6+3-6*5=6+3-30​​​​​​​​​​=-21. So the last digit of -21 is 1​​​. Option (B) is correct.

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The last digit of 12^12 + 13^13 14^14×15^15 =

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