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Intern  Joined: 23 Aug 2011
Posts: 28
What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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13 00:00

Difficulty:   15% (low)

Question Stats: 75% (01:20) correct 25% (01:43) wrong based on 281 sessions

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What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70

Originally posted by kkalyan on 07 Oct 2011, 21:45.
Last edited by Bunuel on 14 Dec 2012, 02:13, edited 2 times in total.
Renamed the topic and edited the question.
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Re: PS: Remainder of 7^(10) divided by 100  [#permalink]

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2
7
thangvietnam wrote:
if we do not hav formular, how do I do?

What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70

The remainder when 7^10 is divided by 100 will be the last two digits of 7^10 (for example 123 divided by 100 yields the remainder of 23, 345 divided by 100 yields the remainder of 45).

$$7^{10}=(7^2)^5=49^5$$ --> the units digit of 49^5 will be 9 (the units digit of 9^even is 1 and the units digit of 9^odd is 9).

So, we have that $$7^{10}=49^5$$ has the units digit of 9, thus the units digit of the remainder must also be 9. Only answer D fits.

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2
1
We have such a pattern
7
7*7=49
7*7*7=343
7*7*7*7=...1
7*7*7*7*7=7
So, the last number repeats every 5th time.
10/4 = 2, and remainder is 2.
we choose 7*7
it means that 7*7*7*7*7*7*7*7*7*7=.......49
We divide by 100, it means .....,49
where 49 is a remainder.
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kkalyan wrote:
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Help End Exam Review Section

What is the remainder, after division by 100, of $$7^{10}$$ ?

1
7
43
49
70

PLZ EXPLAIN THE TRICK IN SOLVING THIS TYPE REMAINDER PROBLEM

Using Binomial theorem, last two digits of an exponent can be found as
7^(10)=7^(2*5)=49^5=(-1+50)^5=(-1)^5+5*(-1)^4*50=-1+50(Just considered last 2-digit of the product)=49

Look for Karishma's blogs. You may find more.

Ans: "D"
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Re: PS: Remainder of 7^(10) divided by 100  [#permalink]

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kkalyan wrote:
What is the remainder, after division by 100, of $$7^{10}$$ ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70

7^4 is 2401 .

So we can write 7^10 as ( 7^4 *7^4 * 7^2) divided by 100 ..

This would give us (1*1*49)/ 100 which would give remainder as 49.
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Re: PS: Remainder of 7^(10) divided by 100  [#permalink]

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if we do not hav formular, how do I do?
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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Hey Bunnel,

this rule is general ? like if XXXXXX9^even the unit digit of the remainder is always 1 and XXXXX9^odd the unit digit of the remainder is always 9 ??

Thanks for your help
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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see wrote:
Hey Bunnel,

this rule is general ? like if XXXXXX9^even the unit digit of the remainder is always 1 and XXXXX9^odd the unit digit of the remainder is always 9 ??

Thanks for your help

The units digit of 9^even is 1 and the units digit of 9^odd is 9.

If the units digit of a number is 1, then the remainder when this number will be divided by 100 will have the units digit of 1, for example 231 divided by 100 gives the reminder of 31.

If the units digit of a number is 9, then the remainder when this number will be divided by 100 will have the units digit of 9, for example 239 divided by 100 gives the reminder of 39.
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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what if the answer choise has another value with units digit 9? how do we need to proceed in that case?
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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Ans:

7^10 can be written as 49^5 which can be written as (49^2)^2. 49 when divided by 100 it will give a remainder of (1)^2.49=49 answer (D).
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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7 ^1 has last digit is 7
7^2 has last digit is 9
3
1

the last digit of 7^10 must be 9

the remainder must has the same last digit

only D fits
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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priyamne wrote:
Ans:

7^10 can be written as 49^5 which can be written as (49^2)^2. 49 when divided by 100 it will give a remainder of (1)^2.49=49 answer (D).

49^5 is not (49^2)^2

Would you please explain yourself. I'm trying to follow your approach but don't quite get it

Thanks
Cheers!
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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Amateur wrote:
what if the answer choise has another value with units digit 9? how do we need to proceed in that case?

Good question. Any clues Bunnuel
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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kkalyan wrote:
What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70

sol:

7=7
7^2=..9
7^3=..3
7^4=1

now 10/4= 2 i.e. second from top of the pattern...which is 9
since we are dividing the number by 100 last number will be reminder
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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jlgdr wrote:
priyamne wrote:
Ans:

7^10 can be written as 49^5 which can be written as (49^2)^2. 49 when divided by 100 it will give a remainder of (1)^2.49=49 answer (D).

49^5 is not (49^2)^2

Would you please explain yourself. I'm trying to follow your approach but don't quite get it

Thanks
Cheers!
J Agree... this approach is incorrect......... $$(49^2)^2 = 49^{(2*2)} = 49^4$$
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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himanshujovi wrote:
Amateur wrote:
what if the answer choise has another value with units digit 9? how do we need to proceed in that case?

Good question. Any clues Bunnuel

When the units digit is 9, the remainder has to be 49. Any other remainder ending with units digit 9 is not possible
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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Bunuel wrote:
thangvietnam wrote:
if we do not hav formular, how do I do?

What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70

The remainder when 7^10 is divided by 100 will be the last two digits of 7^10 (for example 123 divided by 100 yields the remainder of 23, 345 divided by 100 yields the remainder of 45).

$$7^{10}=(7^2)^5=49^5$$ --> the units digit of 49^5 will be 9 (the units digit of 9^even is 1 and the units digit of 9^odd is 9).

So, we have that $$7^{10}=49^5$$ has the units digit of 9, thus the units digit of the remainder must also be 9. Only answer D fits.

excellent. I can not say a word for this wonderful explanation. thank you Buuney
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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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kkalyan wrote:
What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70

7^4 = 2401/100 will have remainder 1
So, 7^8 will have remainder 1

7^10 = 7^8 x 7^2

7^2 will have remainder 49 when divided by 100

So, remainder will be (D) 49

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Re: What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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7^10=(50-1)^2(50-1)^2.49=(50^2-2*50+1)(5^2-2*50+1)*49
for the first two expression, all of number can be divided by 100, only 1*1 is not, so
1*1*49 is the remainder.

hard one.
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What is the remainder, after division by 100, of 7^10 ?  [#permalink]

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kkalyan wrote:
What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70

First note that:

> 1 is the remainder of 101 (=100+1) divided by 100
> 32 is the remainder of 532 (=5*100+32) divided by 100
> 47 is the remainder of 7847 (=78*100+47) divided by 100

$${7^{10}} = K \cdot 100 + R{\mkern 1mu} {\mkern 1mu} \,\,{\mkern 1mu} {\mkern 1mu} \left( {K\,\,{\mathop{\rm int}} \,\,,\,\,\,0 \le R \le 99\,\,{\mathop{\rm int}} } \right){\mkern 1mu}$$

$$? = R$$

$${7^{10}} = {\left( {{7^2}} \right)^5} = {49^5}$$

$${49^2} = {\left( {50 - 1} \right)^2} = {5^2} \cdot {10^2} - 100 + 1 = M \cdot 100 + 1\,\,\,,\,\,\,M\,\,{\mathop{\rm int}} \ge 1\,\,\,\,\,\,\,\,\,\,\left( {M = {5^2} - 1} \right)$$

$${49^4} = {\left( {M \cdot 100 + 1} \right)^2} = {M^2} \cdot {10^4} + M \cdot 200 + 1 = N \cdot 100 + 1\,\,\,,\,\,\,\,N\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\,\,\,\,\,\left( {N = {M^2} \cdot {{10}^2} + 2M} \right)$$

$${49^5} = 49 \cdot \left( {N \cdot 100 + 1} \right) = K \cdot 100 + 49\,\,\,,\,\,\,\,K\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\left( {K = 49N} \right)$$

$$? = 49$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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