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The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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22 Apr 2015, 02:27
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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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22 Apr 2015, 02:53
Bunuel wrote: The last digit of 12^12 + 13^13 – 14^14×15^15 =
A. 0 B. 1 C. 5 D. 8 E. 9
Kudos for a correct solution. This question tests the cyclicity concept. 2 has a cyclicity of 4 i.e. 5th power of 2 has same unit digit as 1st power of 2 . 3 also has cyclicity of 4 4 has cycling of 2 5 has 0. So this complicated calculation boils down to 2^12 +3^13  4^14 ×5^15 6+36×0=9 as unit digit is answer. Answer E
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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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22 Apr 2015, 05:01
Bunuel wrote: The last digit of 12^12 + 13^13 – 14^14×15^15 =
A. 0 B. 1 C. 5 D. 8 E. 9
Kudos for a correct solution. Using cyclicity of 2^12 will have unit digit as 6 (12/4=3 no remainder, 2^4 = 16, so 6) similarly for 3^13 the unit digit is 3 For 4^14 unit digit is 6 For 5^15 unit digit is 5 Now ( 6 + 3  (6*5)) = 9( 30) => 90 (considering only unit digit) = 6



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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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22 Apr 2015, 17:49
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Last digit of:
12^12 is 6 13^13 is 3 14^14 is 6 15^15 is 5
6 + 3  6 x 5 =21
1 is the last digit
Answer: B



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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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27 Apr 2015, 03:09
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Bunuel wrote: The last digit of 12^12 + 13^13 – 14^14×15^15 =
A. 0 B. 1 C. 5 D. 8 E. 9
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Solution: B Units digits of exponents are cyclical, and once you’re determined the full cycle of a particular units digit you can simply extrapolate its pattern forward. Checking these cycles for the units digits of 2, 3, 4, and 5 above: we know that units digits of powers of 2 follow a cycle of 2, 4, 8, 6. Therefore the units digit of 12^12 is 6. The last digits of powers of 3 follow a cycle of 3, 9, 7, 1. Therefore the units digit of 13^13 is 3. The units digits of 4 follow a cycle of 4, 6. Hence the units digit of 14^14 is 6. The units digit of any power of 5 remains 5 so the last digit of 15^15 will be 5. This implies that last digit of 12^12+13^13 will be 9 and that of 14^14×15^15 will be 0. Since 14^14×15^15 is definitely much greater than (12^12+13^13),(12^12+13^13)–(14^14×15^15) will be a negative number with a units digit of 1. Especially if you selected 9 as your answer, remember this little GMAT trick well – for units digit problems that involve subtraction, beware the case in which the second number will be larger than the first, in which case a negative number will result!
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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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09 Jun 2015, 22:14
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mayukh55 wrote: The last digit of 12^12+13^13–14^14×15^15 = a. 0 b. 1 c. 5 d. 8 e. 9 Can anyone explain the solution of this problem ? There are 3 terms here: \(12^{12}\) \(+ 13^{13}\) \(– 14^{14}×15^{15}\)  This term is very simple to manage in case of last digit. An even number multiplied by a multiple of 5 will end in 0. Also, note that it is a huge term as compared with the other two terms. \(12^{12}\)  2 has a cyclicity of 2  4  8  6 so 12 2s will end in 6 \(13^{13}\)  3 has a cyclicity of 3  9  7  1 so 13 3s will end in 3 (A number ending in 6) + (A number ending in 3) gives a number ending in 9. So we have Number ending in 9  Number ending in 0 (much bigger) so the result will end in 1 and will be negative. Answer (B)
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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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13 Feb 2016, 00:41
VeritasPrepKarishma wrote: mayukh55 wrote: The last digit of 12^12+13^13–14^14×15^15 = a. 0 b. 1 c. 5 d. 8 e. 9 Can anyone explain the solution of this problem ? There are 3 terms here: \(12^{12}\) \(+ 13^{13}\) \(– 14^{14}×15^{15}\)  This term is very simple to manage in case of last digit. An even number multiplied by a multiple of 5 will end in 0. Also, note that it is a huge term as compared with the other two terms. \(12^{12}\)  2 has a cyclicity of 2  4  8  6 so 12 2s will end in 6 \(13^{13}\)  3 has a cyclicity of 3  9  7  1 so 13 3s will end in 3 (A number ending in 6) + (A number ending in 3) gives a number ending in 9. So we have Number ending in 9  Number ending in 0 (much bigger) so the result will end in 1 and will be negative. Answer (B) Hi, In order to make this concept more clear, I have a query: If we had Number ending in 9  Number ending in 3 (much bigger), then the answer would have been 4 in negative and not 6. Am I correct ?



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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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13 Feb 2016, 00:52
Devlikes wrote: VeritasPrepKarishma wrote: mayukh55 wrote: The last digit of 12^12+13^13–14^14×15^15 = a. 0 b. 1 c. 5 d. 8 e. 9 Can anyone explain the solution of this problem ? There are 3 terms here: \(12^{12}\) \(+ 13^{13}\) \(– 14^{14}×15^{15}\)  This term is very simple to manage in case of last digit. An even number multiplied by a multiple of 5 will end in 0. Also, note that it is a huge term as compared with the other two terms. \(12^{12}\)  2 has a cyclicity of 2  4  8  6 so 12 2s will end in 6 \(13^{13}\)  3 has a cyclicity of 3  9  7  1 so 13 3s will end in 3 (A number ending in 6) + (A number ending in 3) gives a number ending in 9. So we have Number ending in 9  Number ending in 0 (much bigger) so the result will end in 1 and will be negative. Answer (B) Hi, In order to make this concept more clear, I have a query: If we had Number ending in 9  Number ending in 3 (much bigger), then the answer would have been 4 in negative and not 6. Am I correct ? hi, yes you are right.. and the number need not be much bigger but even just bigger than left side so as to give us a negative number is enough to give 4 as remainder..
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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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31 Jul 2016, 19:03
Devlikes wrote: VeritasPrepKarishma wrote: mayukh55 wrote: The last digit of 12^12+13^13–14^14×15^15 = a. 0 b. 1 c. 5 d. 8 e. 9 Can anyone explain the solution of this problem ? There are 3 terms here: \(12^{12}\) \(+ 13^{13}\) \(– 14^{14}×15^{15}\)  This term is very simple to manage in case of last digit. An even number multiplied by a multiple of 5 will end in 0. Also, note that it is a huge term as compared with the other two terms. \(12^{12}\)  2 has a cyclicity of 2  4  8  6 so 12 2s will end in 6 \(13^{13}\)  3 has a cyclicity of 3  9  7  1 so 13 3s will end in 3 (A number ending in 6) + (A number ending in 3) gives a number ending in 9. So we have Number ending in 9  Number ending in 0 (much bigger) so the result will end in 1 and will be negative. Answer (B) Hi, In order to make this concept more clear, I have a query: If we had Number ending in 9  Number ending in 3 (much bigger), then the answer would have been 4 in negative and not 6. Am I correct ? Just take a number greater than 9, that ends in 3. for ex. 13. So, 139=4. like in 0..take 10, so 109=1
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Re: The last digit of 12^12 + 13^13 – 14^14×15^15 = [#permalink]
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31 Jul 2016, 21:24
Devlikes wrote: VeritasPrepKarishma wrote: mayukh55 wrote: The last digit of 12^12+13^13–14^14×15^15 = a. 0 b. 1 c. 5 d. 8 e. 9 Can anyone explain the solution of this problem ? There are 3 terms here: \(12^{12}\) \(+ 13^{13}\) \(– 14^{14}×15^{15}\)  This term is very simple to manage in case of last digit. An even number multiplied by a multiple of 5 will end in 0. Also, note that it is a huge term as compared with the other two terms. \(12^{12}\)  2 has a cyclicity of 2  4  8  6 so 12 2s will end in 6 \(13^{13}\)  3 has a cyclicity of 3  9  7  1 so 13 3s will end in 3 (A number ending in 6) + (A number ending in 3) gives a number ending in 9. So we have Number ending in 9  Number ending in 0 (much bigger) so the result will end in 1 and will be negative. Answer (B) Hi, In order to make this concept more clear, I have a query: If we had Number ending in 9  Number ending in 3 (much bigger), then the answer would have been 4 in negative and not 6. Am I correct ? How do you subtract one number out of another? The one with the greater absolute value goes on the top and the smaller absolute value goes under it. You subtract and the result gets the sign of the greater absolute value. 100  29 > 100 29  071  29  100 > 100 29  071  But since the sign of 100 is negative, your answer is 71. So the number with greater absolute value is always on top. Number ending in 9  Number ending in 0 (much bigger) will look like> .........0  ......9  ..........1  Answer will be negative ending in 1. Number ending in 9  Number ending in 3 (much bigger) will look like > ............3  .........9  ............4  Answer will be negative ending in 4.
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