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

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.

Answer: D.
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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.

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"

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.

if we do not hav formular, how do I do?

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

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).

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

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

Good question. Any clues Bunnuel

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
check the answers....D is the only choice

Agree... this approach is incorrect......... \((49^2)^2 = 49^{(2*2)} = 49^4\)

When the units digit is 9, the remainder has to be 49. Any other remainder ending with units digit 9 is not possible

excellent. I can not say a word for this wonderful explanation. thank you Buuney

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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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.

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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