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Question

# Find the second right most non-zero digit in the number N=24701613.___

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Solution

## Approach 1: Conventional 24701613=2471613×101613. The second right most non-zero digit in the number would be the digit at the ten's place of the number 2471613. 2471613100=24716134×25 The remainder when 2471613 is divided by 4 is 3. So, 2471613=4k+3. Now we need to find the remainder when 2471613 is divided by 25. 247161325=((250−3)1613)25 R=(−31613)25=((−33)×((310)161))25=(−27×(59049161)25)=((−2)×(−1)16125) Therefore the remainder is 2. So, 2471613=25n+2. 4k+3 =25n+2. For k to be an integer, possible values of n are 1, 5, 9, 13... Corresponding values of k are 6, 31, 56, 81... Therefore, values of k are of the form: 25m -19. 2471613=4(25m−19)+3=100m−73. Therefore, the last two digits is 100-73=27. So, the second right most non-zero digit in the number N=24701613 is 2. Approach 2: Using the techique of last two digits We need to consider only the last two digits. The number can be written as: (47)1613=(474)403×47=(472×472)403×47 =(09×09)403×47=(81)403×47=41×47=....27

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