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# Exponents Problem (m04q20)

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Exponents Problem (m04q20) [#permalink]

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19 Jun 2007, 20:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the last digit of the following number $$2^{22} * 3^{15} * 5^{16} * 7^1$$ ?

(A) 6
(B) 5
(C) 2
(D) 1
(E) 0

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain how to solve these kind of problems

Thanks
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19 Jun 2007, 21:11
1
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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.
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19 Jun 2007, 23:05
2
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sumande wrote:
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.
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19 Jun 2007, 23:30
1
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Thanks for pointing that out !!
But if there is a 2 and a 5, the other numbers don't matter
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21 Jun 2007, 10:24
1
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To be honest, I found this solution after reading yours. But the methodology may be helpful for solving a similar problem :

2^22*3^15*5^16*7^1 = 2*5*(2^21*3^15*5^15*7)
2^22*3^15*5^16*7^1 = 10 * integer

10 * integer ends with 0
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21 Jun 2007, 13:34
1
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2^22)*(3^15)*(5^16)*(7^1)

7^1 = 7 then

5^16 will result 5 in units digit

7 * 5 reults 5 in units digit

Now 5 * 2^anything - will result in 0 as the units digit

Now 0* 3^anything - will give 0 in the units digit.

Answer: Last digist is 0
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21 Jun 2007, 19:54
I found this another way as well. However, I think the method used in the second post is the best one to follow:

(2^15)(3^15)(5^16)(7^1)
=(6^15)(5^16)(7^1)
=(6^15)(5^15)(5^1)(7^1)
=(30^15)(35^1)

30^x = the last digit will always = 0
35 * 30 = the last digit will always = 0
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Re: Exponents Problem (m04q20) [#permalink]

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12 Oct 2010, 05:14
We have 2 and 5 in the product, so last digit always be 0 irrespective of any other number.
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Re: Exponents Problem (m04q20) [#permalink]

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12 Oct 2010, 22:54
trahul4 wrote:
What is the last digit of the following number $$2^{22} * 3^{15} * 5^{16} * 7^1$$ ?

(A) 6
(B) 5
(C) 2
(D) 1
(E) 0

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain how to solve these kind of problems

Thanks

E

2() * 3() * 5() * 7() = multiple of 10 ==> 0
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Re: Exponents Problem (m04q20) [#permalink]

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13 Oct 2010, 09:03
4
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The first thing i saw in this question was a presence of a 5. That means the answer had to have either a 0 or a 5.

So now I have automatically eliminated 3 answers.

Second, all powers of 3 result in an odd number. (Check it up: 3^1=3, 3^4=81).

Third, all powers of 2 are even.

So its basically even (because of 2) mulitplied by odd (because of 3)

even*odd = even

even number*(any power of 5) = ends with a zero
eg. 2*5=10, 2*25=50, etc.

Hence E
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Re: Exponents Problem (m04q20) [#permalink]

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13 Oct 2010, 17:54
Look no further if you see 2 and 5 as one of the factors.
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Re: Exponents Problem (m04q20) [#permalink]

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14 Oct 2010, 07:16
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
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Re: Exponents Problem (m04q20) [#permalink]

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14 Oct 2010, 20:43
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
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Re: Exponents Problem (m04q20) [#permalink]

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14 Oct 2011, 05:29
i also got the ans..which will be 4*7*5*7 which will make the unit digit 0
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Re: Exponents Problem (m04q20) [#permalink]

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14 Oct 2011, 12:44
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
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Re: Exponents Problem (m04q20) [#permalink]

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15 Oct 2011, 04:14
zero..

any 5 when multiplied by 2 ll result a zero as its unit digit.
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Re: Exponents Problem (m04q20) [#permalink]

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17 Oct 2011, 03:40
trahul4 wrote:
What is the last digit of the following number $$2^{22} * 3^{15} * 5^{16} * 7^1$$ ?

(A) 6
(B) 5
(C) 2
(D) 1
(E) 0

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

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

Hence
2^22 = 2^(22/4) = 2^2 = 4
3^15 = 3^(15/4) = 3^1 = 3
5^16 = 5^1 = 5
7^1 = 7

4*3*5*7 = 420 .. untis digit 0
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Re: Exponents Problem (m04q20) [#permalink]

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17 Nov 2011, 23:01
2- 2,4,8,6
3 - 3,9,7,1
5- always 5

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.

So 4x3x7x5x7 = 0

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Re: Exponents Problem (m04q20) [#permalink]

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16 Oct 2012, 07:01
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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!!!!
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Re: Exponents Problem (m04q20) [#permalink]

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15 Oct 2013, 08:17
E.

2 ways to do this one.

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

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# Exponents Problem (m04q20)

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