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We need to find the last digit of each term first.

2^1=2; 2^2=4; 2^3=8; 2^4=16; 2^5=32.
So the last digit of 2^x is repeated every 4th term.
so, the last digit of 2^22 will be the last digit of 2^2=4.

Similarly, for 3^x the last digit is repeated every 4th term.
3^1=3; 3^2=9; 3^3=27; 3^4=81; 3^5=243
so, the last digit of 3^15 will be last digit of 3^3=7

The last digit of 5^x is always 5 and that of 7^1 is 7.

So the last digit of the product will be the last digit of 4*3*5*7=0.

We need to find the last digit of each term first.

2^1=2; 2^2=4; 2^3=8; 2^4=16; 2^5=32. So the last digit of 2^x is repeated every 4th term. so, the last digit of 2^22 will be the last digit of 2^2=4.

Similarly, for 3^x the last digit is repeated every 4th term. 3^1=3; 3^2=9; 3^3=27; 3^4=81; 3^5=243 so, the last digit of 3^15 will be last digit of 3^3=7

The last digit of 5^x is always 5 and that of 7^1 is 7.

So the last digit of the product will be the last digit of 4*3*5*7=0.

the process is perfect and result too but a small mistake: 4 x 7 x 5 x 7 = 0.

take the last digit of all of them and t hen multiply and find 2^0=1;2^1=2;2^2=4;2^3=8;2^4=6;2^5=2;so it repeats after 4 times 3^0=1;3^1=3;3^2=9;3^3=7;3^4=1;3^5=3;so it repeats after 4 times 5^0=1;5^1=2;5^2=5;so it repeats after 1 times 7^1=7

2^22=22/4=2 is remainder.means it should repeat after 2 times.i.e 8 3^15=15/4=3 is remainder.means it should repeat after 3 times.i.e 7 5 and 7 so total=8*7*5*7=last digit is 0

2^x will repeat the last digit 2 after each 4th power.So 2^22=(2^4)^5 * 2^2 = yyy2*4=yyy8 I am just considering the last digit. now 3^4=y1 3^15= (3^4)^3 * 3^3 = yy1*27 =yyy7 5^16=yyy5 7^1=7 Multiplying all last digit=8*7*5*7=yy0 so last digit is 0.Ans must be E Consider KUDOS if u like the answer.

This was easier than I thought. I did no math. This really wasn't an exponent question but a number properties question:

Since 2 is a factor the answer cannot be odd so B and D are out Since 5 is a factor and all multiples of 5 are end in 5 or 0 then the answer is E. 0 _________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell

Can someone please explain how to solve these kind of problems

Thanks

This can be solved by identifying the patterns for 2 : 2*1 = 2 , 2*2 = 4 , 2^3 = 8 , 2^4 = 16 hence it follows : 2,4,8,6 -- 4 is the count similarly for 3 : 3,9,7,1 -- 4 for 5 : 5 -- count is 1 for 7 : 7,9,3,1 -- count is 4

So we know that 2^22 will end in a 4 (goes through five rotations and then two more places), 3 will end on a 7 (3 rotations then 3 more places), 5 is 5, and 7 is 7.

2*7*5*7 --- i like this thought process i would have never thought of it this way! i would have seen the 2 and 5 and thought oh it's 0 but now when i'm faced with a question like this and there is no 2 or 5 i know what to do!!!!

Longer way: We only care about the units digit ("ones" place - from here on out abbreviated as UD) of each term. So we need to figure out what the UD is of each of the terms. The best way to do this is to look for a repeating pattern.

For example, the pattern for units digit for 2 raised to a power is: \(2^1\) = 2 \(2^2\) = 4 \(2^3\) = 8 \(2^4\) = 16 (units digit is 6) \(2^5\) = 32 (units digit is 2 - here the pattern begins to repeat)

So, the 2, 4, 8, 6 pattern repeats every 4 powers.

Next step to find the units digit of 2^22, we look at 22. 22 divided by 4 = 5 remainder 2. If the pattern is 2, 4, 8, 6, we need to count to the second term (remember, the remainder was 2!)...the second term in the patter is 4. So the units digit is 4.

We repeat the process for 3^15 (UD is 3), 5^16 (UD is 5), and 7^1 (UD is 7).

Finally, we multiply all the UDs together: 4 x 3 x 5 x 7 = 420. The UD of 420 is 0. So, this is the answer.

Short cut:

Since a 2 and a 5 are in the term, the UD will always be 0, no matter what else is in the term (2x5=10).

gmatclubot

Re: Exponents Problem (m04q20)
[#permalink]
15 Oct 2013, 08:17