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When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?
- a)24
- b)124
- c)18
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
When 7179 and 9699 are divided by another natural number N , remainder...
We are given that when 7179 and 9699 are divided by a natural number N, the remainder obtained is the same. This means that 7179 - 9699 must be divisible by N.
First, calculate the difference between 9699 and 7179:
9699 - 7179 = 2520
Now, we need to find the divisors of 2520 that end with one or more zeros. ie multiples of 10.
Step 1: Find the divisors of 2520.
2520 = 2^3 × 3^2 × 5 × 7
l number of divisors is given by the product of one plus each of the exponents in the prime factorization:
(3 + 1) × (2 + 1) × (1 + 1) × (1 + 1) = 4 × 3 × 2 × 2 = 48
So, 2520 has 48 divisors in total.
Step 2: Find divisors of 2520 that are multiples of 10
.To be a multiple of 10, the divisor must include at least one factor of 2 and one factor of 5.
From the prime factorization of 2520, we know that there are 3 factors of 2 and 1 factor of 5, so the divisors of 2520 that are multiples of 10 must include at least one 2 and one 5.
the remaining factors after we take out one factor of 2 and one factor of 5 from the prime factorization of 2520:
2520 / 10 = 2^2 × 3^2 × 7
The number of divisors of 2^2 × 3^2 × 7 is:
(2 + 1) × (2 + 1) × (1 + 1) = 3 × 3 × 2 = 18
So, there are 18 divisors of 2520 that are multiples of 10.
Step 3: Conclusion The number of values of N that end with one or more zeros is 18.
Thus, the correct answer is:
c) 18.
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Community Answer
When 7179 and 9699 are divided by another natural number N , remainder...
Understanding the Problem
To solve the problem, we need to find a natural number N such that when both 7179 and 9699 are divided by N, the remainder is the same. This condition implies that the difference between the two numbers should be divisible by N.
Calculating the Difference
- First, calculate the difference:
- 9699 - 7179 = 2520
We need to find all divisors of 2520 that end with one or more zeros.
Finding Divisors of 2520
- The next step is to determine the divisors of 2520. First, we find its prime factorization:
- 2520 = 2^3 × 3^2 × 5^1 × 7^1
Using this factorization, we can find the total number of divisors using the formula (e1 + 1)(e2 + 1)(e3 + 1)(e4 + 1), where e1, e2, e3, and e4 are the powers of the prime factors.
- For 2520:
- (3 + 1)(2 + 1)(1 + 1)(1 + 1) = 4 × 3 × 2 × 2 = 48
Thus, 2520 has 48 divisors in total.
Identifying Divisors Ending with Zeros
- To find the divisors of 2520 that end with one or more zeros, we focus on those that are multiples of 10. A number ends with at least one zero if it is divisible by 10, which means it needs at least one factor of 2 and one factor of 5.
- The divisors must be of the form: 2^a × 3^b × 5^c × 7^d where:
- a ≥ 1, c ≥ 1 (to ensure divisibility by 10)
- Possible values for a, b, c, and d are:
- a can be 1, 2, or 3 (3 options)
- b can be 0, 1, or 2 (3 options)
- c must be 1 (1 option)
- d can be 0 or 1 (2 options)
Calculating the total combinations:
- Total = 3 × 3 × 1 × 2 = 18
Conclusion
Thus, the number of divisors of 2520 that end with one or more zeros is 18. Therefore, the correct answer is option 'C'.
To solve the problem, we need to find a natural number N such that when both 7179 and 9699 are divided by N, the remainder is the same. This condition implies that the difference between the two numbers should be divisible by N.
Calculating the Difference
- First, calculate the difference:
- 9699 - 7179 = 2520
We need to find all divisors of 2520 that end with one or more zeros.
Finding Divisors of 2520
- The next step is to determine the divisors of 2520. First, we find its prime factorization:
- 2520 = 2^3 × 3^2 × 5^1 × 7^1
Using this factorization, we can find the total number of divisors using the formula (e1 + 1)(e2 + 1)(e3 + 1)(e4 + 1), where e1, e2, e3, and e4 are the powers of the prime factors.
- For 2520:
- (3 + 1)(2 + 1)(1 + 1)(1 + 1) = 4 × 3 × 2 × 2 = 48
Thus, 2520 has 48 divisors in total.
Identifying Divisors Ending with Zeros
- To find the divisors of 2520 that end with one or more zeros, we focus on those that are multiples of 10. A number ends with at least one zero if it is divisible by 10, which means it needs at least one factor of 2 and one factor of 5.
- The divisors must be of the form: 2^a × 3^b × 5^c × 7^d where:
- a ≥ 1, c ≥ 1 (to ensure divisibility by 10)
- Possible values for a, b, c, and d are:
- a can be 1, 2, or 3 (3 options)
- b can be 0, 1, or 2 (3 options)
- c must be 1 (1 option)
- d can be 0 or 1 (2 options)
Calculating the total combinations:
- Total = 3 × 3 × 1 × 2 = 18
Conclusion
Thus, the number of divisors of 2520 that end with one or more zeros is 18. Therefore, the correct answer is option 'C'.
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