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Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.
- a)6
- b)8
- c)9
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Raju had to divide 1080 by N, a two-digit number. Instead, he performe...
According to the question,
Given information is,Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M, which is obtained by reversing the digits of N, and ended up with a quotient that was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.
1080 = 23 x 32 x 51.
Now, N x Q = 1080 and M x (Q - 25) = 1080
By Doing hit and trial method,
N = 27
Hence the answer is 9.
Now, N x Q = 1080 and M x (Q - 25) = 1080
By Doing hit and trial method,
N = 27
Hence the answer is 9.
Most Upvoted Answer
Raju had to divide 1080 by N, a two-digit number. Instead, he performe...
Option C - 9 Any number whose sum of its digits is a multiple of 9 is divisible by 9
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Raju had to divide 1080 by N, a two-digit number. Instead, he performe...
Let's assume that the two-digit number N is a two-digit number AB, where A and B are digits.
Finding M:
To obtain M, Raju reverses the digits of N. Therefore, M is a two-digit number BA.
Using M instead of N:
Raju performs the division using M instead of N. Let's assume that the quotient he obtains while using M is Q.
Using N:
If Raju had used N instead of M, he would have obtained a quotient of 1080/N.
Given that the quotient he obtained using M is 25 less than what he should have obtained using N:
Q = 1080/N - 25
Both N and M divide 1080:
N and M are factors of 1080, which means that they divide 1080 without leaving any remainder.
1080/N and 1080/M are integers:
Since N and M divide 1080 without leaving any remainder, the quotients 1080/N and 1080/M are integers.
Using this information, we can obtain some constraints on the possible values of N and M.
Possible values of AB:
Since N is a two-digit number, A cannot be 0. Therefore, A can be any of the digits from 1 to 9.
B can also be any of the digits from 1 to 9. However, we can narrow down the possible values of B by considering the fact that 1080/N and 1080/M are integers.
Constraints on B:
If B is odd, then the last digit of 1080/B will be 0 or 5. Therefore, B cannot be an odd number.
If B is even, then the last digit of 1080/B will be 0, 2, 4, 6 or 8. Therefore, B can be any even number from 2 to 8.
Possible values of BA:
Using the constraints on the possible values of A and B, we can obtain the possible values of BA (i.e., M).
BA can be any of the numbers from 12 to 98, excluding the odd numbers and the numbers that end in 0.
For example, BA can be any of the following numbers: 24, 26, 28, 42, 46, 48, 62, 64, 68, 82, 84, 86.
Finding N:
We know that Q = 1080/N - 25. We also know that 1080/N is an integer. Therefore, Q must be an integer.
Since Q is 25 less than 1080/N, we can write:
1080/N - Q = 25
We can rearrange this equation as follows:
N = 1080/(Q+25)
We know that N is a two-digit number AB. Therefore, we can write:
N = 10A + B
We can substitute this expression for N in the equation above and simplify:
10A + B = 1080/(Q+25)
10A(Q+25) + B(Q+25) = 1080
10AQ + 10A*25 + BQ + 25B = 1080
10AQ + BQ + 25A + 25B = 1080
Q
Finding M:
To obtain M, Raju reverses the digits of N. Therefore, M is a two-digit number BA.
Using M instead of N:
Raju performs the division using M instead of N. Let's assume that the quotient he obtains while using M is Q.
Using N:
If Raju had used N instead of M, he would have obtained a quotient of 1080/N.
Given that the quotient he obtained using M is 25 less than what he should have obtained using N:
Q = 1080/N - 25
Both N and M divide 1080:
N and M are factors of 1080, which means that they divide 1080 without leaving any remainder.
1080/N and 1080/M are integers:
Since N and M divide 1080 without leaving any remainder, the quotients 1080/N and 1080/M are integers.
Using this information, we can obtain some constraints on the possible values of N and M.
Possible values of AB:
Since N is a two-digit number, A cannot be 0. Therefore, A can be any of the digits from 1 to 9.
B can also be any of the digits from 1 to 9. However, we can narrow down the possible values of B by considering the fact that 1080/N and 1080/M are integers.
Constraints on B:
If B is odd, then the last digit of 1080/B will be 0 or 5. Therefore, B cannot be an odd number.
If B is even, then the last digit of 1080/B will be 0, 2, 4, 6 or 8. Therefore, B can be any even number from 2 to 8.
Possible values of BA:
Using the constraints on the possible values of A and B, we can obtain the possible values of BA (i.e., M).
BA can be any of the numbers from 12 to 98, excluding the odd numbers and the numbers that end in 0.
For example, BA can be any of the following numbers: 24, 26, 28, 42, 46, 48, 62, 64, 68, 82, 84, 86.
Finding N:
We know that Q = 1080/N - 25. We also know that 1080/N is an integer. Therefore, Q must be an integer.
Since Q is 25 less than 1080/N, we can write:
1080/N - Q = 25
We can rearrange this equation as follows:
N = 1080/(Q+25)
We know that N is a two-digit number AB. Therefore, we can write:
N = 10A + B
We can substitute this expression for N in the equation above and simplify:
10A + B = 1080/(Q+25)
10A(Q+25) + B(Q+25) = 1080
10AQ + 10A*25 + BQ + 25B = 1080
10AQ + BQ + 25A + 25B = 1080
Q
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Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.a)6b)8c)9d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.a)6b)8c)9d)None of theseCorrect answer is option 'C'. Can you explain this answer? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.a)6b)8c)9d)None of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.a)6b)8c)9d)None of theseCorrect answer is option 'C'. Can you explain this answer?.
Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.a)6b)8c)9d)None of theseCorrect answer is option 'C'. Can you explain this answer? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.a)6b)8c)9d)None of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Raju had to divide 1080 by N, a two-digit number. Instead, he performed the division using M which is obtained by reversing the digits of N and ended up with a quotient which was 25 less than what he should have obtained otherwise. If 1080 is exactly divisible both by N and M, find the sum of the digits of N.a)6b)8c)9d)None of theseCorrect answer is option 'C'. Can you explain this answer?.
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