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The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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03 Jan 2013, 05:28
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LAST TWO DIGITS OF A NUMBER Question:The last two digits of (31^786)(19^266)(2^101) is A. 14 B. 22 C. 36 D. 72 E. 92 Lets check how to drill this. We will discuss the last two digits of numbers ending with the following digits in sets: a) 1 b) 3,7&9 c) 2, 4,6&8 d) 5 a) Number ending with 1: Ex : Find the last 2 digits of 31^786 Now, multiply the 10s digit of the number with the last digit of exponent 31^78 6= 3*6=1 8 ==> 8 is the 10s digit. Units digit is obviously 1 So, last 2 digits are => 81 b) Number ending with 3 ,7 & 9: Ex: Find last 2 digits of 19^266 We need to get this in such as way that the base has last digit as 1 19^266 = (19^2)^133 = 3 61^13 3Now, follow the previous method => 6 * 3 = 18 = 18 8 is 10s digit. So, last two digits are => 81 Remember: 3^4=81 7^4=2401 9^2 = 81 Ex 2: Find last two digits of 33^288 Now, 33^288 = (33^4)^72 = (xx 21 )^7 2Tens digit is > 2*2 = 04 > 4 So, last two digits are => 41 Ex 3: find last 2 digits of 87^474 (87^2)*(87^4)^118 => (xx69) * (xx 61 )^11 8 ==> (6 x 8 = 48 ==> 8 is 10s digit) => (xx69)*(81) So, last two digits are 89 c)Ending with 2, 4, 6 or 8: Here, we use the fact that 76 power any number gives 76. i.e 76^n = ...xxx76 (e.g. 76^2 = 5,776, 76^3 = 438,976) We also need to remember that, 24^2 = xx76 2^10 xx24 24^even = xx76 24^odd = xx24 Ex: Find the last two digits of 2^543 2^543 = ((2^10)^54) * (2^3) = ((xx24)54)* 8 = ((xx76)^27)*8 76 power any number is 76 Which gives last digits as => 76 * 8 = 608 So last two digits are : 08 c)Ending with 5: 5 has its own special cases: The tens digit of a number raised to a power changes based on tens digits of the number itself. Say, we have following number: 05, 15, 25, 35… say> x = 05 x^2 = 25x^3 = 1 25x^4 = 6 25as the tens digit of x is 0(i.e. even), last two digits has to be 25. x = 15 x^2 = 2 25x^3 = 33 75x^4 = 506 25x^5 = 7593 75as the tens digit of x is 1(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on. x = 25 x^2 = 6 25x^3 = ..XX 25x^4 = ..XX 25x^5 = ..XX 25as the tens digit of x is 0(i.e. even), last two digits has to be 25. x = 35 x^2 = 12 25x^3 = ..XX 75x^4 = ..XX 25x^5 = ..XX 75as the tens digit of x is 3(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on From above, we see that (1) if number's 10s digit is even, then last digit will always has to be 25. (2) if number's 10s digit is odd, then last digit will always has to be in cyclic pattern of 25, 75,25,75 and so on... Now, again, try the question posted on the top.
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Last edited by goutamread on 03 Jan 2013, 11:17, edited 1 time in total.



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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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03 Jan 2013, 10:07
Very laborious isnt it?
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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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03 Jan 2013, 10:19
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Question:The last two digits of (31^786)(19^266)(2^101) is A. 14 B. 22 C. 36 D. 72 E. 92 Solution: 31^786==> Unit Digits: as last digit is 1, units digit has to be 1. Tens digit: 31^78 6 ==> as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=1 8 and thus, 8 is the tens digit. Thus, last two digit of expression 31^786 = 81 19^266 ==> Unit Digits: as last digit is 9, as we know pattern power raise to 9: 9,1,9,1... i.e. it repeats after every 2 iterations.. as 266 is even i.e. dividible by 2, we cud say,units digit has to be 1. Tens digit: 19^266 = (19^2)^133 = (xx 61)^13 3 ==> now, as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=1 8 and thus, 8 is the tens digit. Thus, last two digit of expression 19^266 = 81 (2^101) ==> Tens Digit: 2^101 = 2^100 * 2 =[(2^10)^10 ]*2 = [XX24^10]*2 ==> As we know xx24^even = xx76 ==> [xx76 ^ 10]*2 ==> as we knoe anything raised to xx76 is xx76 ==> xx76*2 = XX52 THUS XX81 * XX81 *XX52 = XX 72
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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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03 Jan 2013, 10:27
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rajathpanta wrote: Very laborious isnt it? Rajath  This example was just to elaborate all possible scenarios i.e. (a)1 (b)3,7,9 and (c)2,4,6,8. Indeed, I have detailed each step over here. Once you are used to the concept/rules, this problem shall not take more than the (per question) average time.
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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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03 Jan 2013, 10:38
Interestin methodology. Thanks for the solution



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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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03 Jan 2013, 10:55
[quote="goutamread"]Question:The last two digits of (31^786)(19^266)(2^101) is A. 14 B. 22 C. 36 D. 72 E. 92
Solution: 31^786==> Unit Digits: as last digit is 1, units digit has to be 1. Tens digit: 31^786 ==> as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=18 and thus, 8 is the tens digit. Thus, last two digit of expression 31^786 = 81
19^266 ==> Unit Digits: as last digit is 9, as we know pattern power raise to 9: 9,1,9,1... i.e. it repeats after every 2 iterations.. as 266 is even i.e. dividible by 2, we cud say,units digit has to be 1. Tens digit: 19^266 = (19^2)^133 = (xx61)^133 ==> now, as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=18 and thus, 8 is the tens digit. Thus, last two digit of expression 19^266 = 81
(2^101) ==> Tens Digit: 2^101 = 2^100 * 2 =[(2^10)^10 ]*2 = [XX24^10]*2 ==> As we know xx24^even = xx76 ==> [xx76 ^ 10]*2 ==> as we knoe anything raised to xx76 is xx76 ==> xx76*2 = XX52
THUS XX81 * XX81 *XX52 = XX72[/quotew
what would be second last digit if last digit is 5



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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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03 Jan 2013, 11:16
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chiknichameleon wrote: what would be second last digit if last digit is 5 Actually, 5 has its own special cases: The tens digit of a number raised to a power changes based on tens digits of the number itself. Say, we have following number: 05, 15, 25, 35… say> x = 05 x^2 = 25x^3 = 1 25x^4 = 6 25as the tens digit of x is 0(i.e. even), last two digits has to be 25. x = 15 x^2 = 2 25x^3 = 33 75x^4 = 506 25x^5 = 7593 75as the tens digit of x is 1(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on. x = 25 x^2 = 6 25x^3 = ..XX 25x^4 = ..XX 25x^5 = ..XX 25as the tens digit of x is 0(i.e. even), last two digits has to be 25. x = 35 x^2 = 12 25x^3 = ..XX 75x^4 = ..XX 25x^5 = ..XX 75as the tens digit of x is 3(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on From above, we see that (1) if number's 10s digit is even, then last digit will always has to be 25. (2) if number's 10s digit is odd, then last digit will always has to be in cyclic pattern of 25, 75,25,75 and so on... P.S. Thanks for asking the question. I missed that part. I shall update the post to include that part.
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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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05 Jan 2013, 07:01
but during the gmat exam we are always against time..can that question come on gmat exam or maybe there is another way of answering the question?
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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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31 Aug 2015, 01:56
How will we solve this question  Q. What will be the last two digits in 1122^1122! ? (asked in TCS placement Paper).



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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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21 Sep 2015, 19:22
this is really a good way. my way can lead to the answer but definitely takes way longer time. Thanks a lot, bro.



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Re: The last two digits of (31^786)(19^266)(2^101) is [#permalink]
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10 Apr 2016, 06:33
goutamread wrote: LAST TWO DIGITS OF A NUMBER Question:The last two digits of (31^786)(19^266)(2^101) is A. 14 B. 22 C. 36 D. 72 E. 92 Lets check how to drill this. We will discuss the last two digits of numbers ending with the following digits in sets: a) 1 b) 3,7&9 c) 2, 4,6&8 d) 5 a) Number ending with 1: Ex : Find the last 2 digits of 31^786 Now, multiply the 10s digit of the number with the last digit of exponent 31^78 6= 3*6=1 8 ==> 8 is the 10s digit. Units digit is obviously 1 So, last 2 digits are => 81 b) Number ending with 3 ,7 & 9: Ex: Find last 2 digits of 19^266 We need to get this in such as way that the base has last digit as 1 19^266 = (19^2)^133 = 3 61^13 3Now, follow the previous method => 6 * 3 = 18 = 18 8 is 10s digit. So, last two digits are => 81 Remember: 3^4=81 7^4=2401 9^2 = 81 Ex 2: Find last two digits of 33^288 Now, 33^288 = (33^4)^72 = (xx 21 )^7 2Tens digit is > 2*2 = 04 > 4 So, last two digits are => 41 Ex 3: find last 2 digits of 87^474 (87^2)*(87^4)^118 => (xx69) * (xx 61 )^11 8 ==> (6 x 8 = 48 ==> 8 is 10s digit) => (xx69)*(81) So, last two digits are 89 c)Ending with 2, 4, 6 or 8: Here, we use the fact that 76 power any number gives 76. i.e 76^n = ...xxx76 (e.g. 76^2 = 5,776, 76^3 = 438,976) We also need to remember that, 24^2 = xx76 2^10 xx24 24^even = xx76 24^odd = xx24 Ex: Find the last two digits of 2^543 2^543 = ((2^10)^54) * (2^3) = ((xx24)54)* 8 = ((xx76)^27)*8 76 power any number is 76 Which gives last digits as => 76 * 8 = 608 So last two digits are : 08 c)Ending with 5: 5 has its own special cases: The tens digit of a number raised to a power changes based on tens digits of the number itself. Say, we have following number: 05, 15, 25, 35… say> x = 05 x^2 = 25x^3 = 1 25x^4 = 6 25as the tens digit of x is 0(i.e. even), last two digits has to be 25. x = 15 x^2 = 2 25x^3 = 33 75x^4 = 506 25x^5 = 7593 75as the tens digit of x is 1(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on. x = 25 x^2 = 6 25x^3 = ..XX 25x^4 = ..XX 25x^5 = ..XX 25as the tens digit of x is 0(i.e. even), last two digits has to be 25. x = 35 x^2 = 12 25x^3 = ..XX 75x^4 = ..XX 25x^5 = ..XX 75as the tens digit of x is 3(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on From above, we see that (1) if number's 10s digit is even, then last digit will always has to be 25. (2) if number's 10s digit is odd, then last digit will always has to be in cyclic pattern of 25, 75,25,75 and so on... Now, again, try the question posted on the top. The answer is 72 31^786 = (30+1)^786 use binomial theorem The last two digits will be 786*30 + 1 = 81 19^266 = 61^133 133*60+1 = 81 2^101=(2^10)^10*2=24^10*2=76^5*2=76*2=52(the last two digits of any power of 76 is always 76) 81*81*52=61*52=72




Re: The last two digits of (31^786)(19^266)(2^101) is
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