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Five three digit numbers including N,were to be added. While adding,the reverse of N was added by mistake instead of N. Hence, the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit?
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Five three digit numbers including N,were to be added. While adding,th...
Problem Statement: Five three digit numbers including N, were to be added. While adding, the reverse of N was added by mistake instead of N. Hence, the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit.

Solution:
Let the three-digit number N be represented as ABC, where A, B, C are the hundreds, tens, and units digits, respectively.

Step 1: Express the given conditions mathematically.

The sum of the five three-digit numbers, including N, is:
ABC + DEF + GHI + JKL + LKJ = 1000A + 100B + 10C + 100D + 10E + F + 100G + 10H + I + 100J + 10K + L + 100L + 10K + J = 1111(A + D + G + J + L) + 101(B + E + H + K) + 11(C + F + I)

When the reverse of N, which is CBA, is added instead of N, the sum becomes:
CBA + DEF + GHI + JKL + LKJ = 1000C + 100B + 10A + 100D + 10E + F + 100G + 10H + I + 100J + 10K + L + 100L + 10K + J = 1111(C + D + G + J + L) + 101(B + E + H + K) + 11(A + F + I)

According to the problem, the sum increased by 11 times the sum of the digits of N. Therefore, we have:
11(A + B + C) = 1111(C + D + G + J + L) + 101(B + E + H + K) + 11(A + F + I) - 1111(A + D + G + J + L) - 101(B + E + H + K) - 11(C + F + I)

Simplifying the above equation, we get:
10A - 100C = 1100L - 1100D + 100J - 100G + 10I - 10F + 11C - 11A

Also, it is given that:
8(C - A) = 2A + 6

Step 2: Solve the equations to find the value of B.

Using the second equation, we can express A in terms of C as:
A = (4C - 3)/5

Substituting this value of A in the first equation, we get:
8C - 10B - 108 = -100D + 100G - 10I + 10F + 112L

Simplifying the above equation, we get:
4C - 5B - 54 = -50D + 50G - 5I + 5F + 56L

Using the fact that A, B, C are digits, we know that 0 ≤ A ≤ 9, 0 ≤ B ≤ 9, and 1 ≤ C
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Five three digit numbers including N,were to be added. While adding,the reverse of N was added by mistake instead of N. Hence, the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit?
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Five three digit numbers including N,were to be added. While adding,the reverse of N was added by mistake instead of N. Hence, the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about Five three digit numbers including N,were to be added. While adding,the reverse of N was added by mistake instead of N. Hence, the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Five three digit numbers including N,were to be added. While adding,the reverse of N was added by mistake instead of N. Hence, the sum increased by 11 times the sum of the digits of N. Eight times the difference of N's units and hundreds digits is 6 more than twice its hundreds digit. Find its tens digit?.
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