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What is the greatest number which divides 742 and 1162 leaves the remainder of exactly 7 in each case?
Let the greatest number be 'x'.
Then nx + 7 = 742 : nx = 735
And mx + 7 = 1162 :mx = 1155
The number required is the Greatest Common Factor of 735 and 1155.
735 expressed as the product of its prime factors = 3 x 5 x 7 x 7
1155 expressed as the product of its prime factors = 3 x 5 x 7 x 11
The GCF is 3 x 5 x 7 = 105
What else can I help you with?
Find the greatest number which divides 742 and 1162 leaves a remainder of exactly 7 in each case?
This is a slight twist to the normal find the GCF of two numbers. In this case as a remainder of 7 is required, subtracting 7 from each number and then finding the GCF of the resulting numbers will solve the problem: 742 - 7 = 735 1162 - 7 = 1155 GCF of 1155 and 735 (using Euclid's method): 1155 / 735 = 1 r 420 735 / 420 = 1 r 315 420 / 315 = 1 r 105 315 / 105 = 3 r 0 GCF of 735 & 1155 is 105, thus 105 is the greatest number that will divide 742 and 1162 leaving a remainder of exactly 7 each time.
What goes into 231?
To determine factors of a number it is necessary to divide the number by prime numbers of increasing value. If there is no remainder then the prime number is a factor. The square root of 231 is approximately 15 so it is necessary to divide by all the prime numbers up to 13 (15 is not a prime number). 231 ÷ 2 leaves a remainder of 1 231 ÷ 3 divides exactly 77 times (which is 7 x 11 - both are prime numbers) 231 ÷ 5 leaves a remainder of 1 231 ÷ 7 divides exactly 33 times (which is 3 x 11 - both are prime numbers) 231 ÷ 11 divides exactly 21 times (which is 3 x 7 - both are prime numbers) 231 ÷ 13 leaves a remainder of 10 If we multiply together the three numbers that are shown to be factors 3 x 7 x 11 = 231 this confirms that there are no other factors of 231 except 1 and 231 itself. This last condition is true for every positive integer. 231 is divisible by 1, 3, 7, 11 and 231
Find the greatest number which divides 6168 2447 and 3118 leaving the same remainder in each case?
Well, honey, the greatest number that fits the bill is the difference between the numbers. So, 6168 - 2447 = 3721, and 3118 - 2447 = 671. The greatest number that divides all three and leaves the same remainder is the greatest common divisor of 3721 and 671, which is 671.
What is the remainder of 26 divided by 5?
5.2
What number is between 12 and 32 when divided by 5 it leaves a remainder of 1 when divided by 4 it leaves a remainder of 2?
26
What is the number less than seventy-five that divides exactly by seven but leaves remainder one when divided by eleven?
56
Find the greatest number which divides 742 and 1162 leaves a remainder of exactly 7 in each case?
This is a slight twist to the normal find the GCF of two numbers. In this case as a remainder of 7 is required, subtracting 7 from each number and then finding the GCF of the resulting numbers will solve the problem: 742 - 7 = 735 1162 - 7 = 1155 GCF of 1155 and 735 (using Euclid's method): 1155 / 735 = 1 r 420 735 / 420 = 1 r 315 420 / 315 = 1 r 105 315 / 105 = 3 r 0 GCF of 735 & 1155 is 105, thus 105 is the greatest number that will divide 742 and 1162 leaving a remainder of exactly 7 each time.
Is 5 a factor of 42?
No, 5 is not a factor of 42. A factor of a number divides it evenly with no remainder. When you divide 42 by 5, it does not divide evenly and leaves a remainder of 2. Therefore, 5 is not a factor of 42.
What goes into 231?
To determine factors of a number it is necessary to divide the number by prime numbers of increasing value. If there is no remainder then the prime number is a factor. The square root of 231 is approximately 15 so it is necessary to divide by all the prime numbers up to 13 (15 is not a prime number). 231 ÷ 2 leaves a remainder of 1 231 ÷ 3 divides exactly 77 times (which is 7 x 11 - both are prime numbers) 231 ÷ 5 leaves a remainder of 1 231 ÷ 7 divides exactly 33 times (which is 3 x 11 - both are prime numbers) 231 ÷ 11 divides exactly 21 times (which is 3 x 7 - both are prime numbers) 231 ÷ 13 leaves a remainder of 10 If we multiply together the three numbers that are shown to be factors 3 x 7 x 11 = 231 this confirms that there are no other factors of 231 except 1 and 231 itself. This last condition is true for every positive integer. 231 is divisible by 1, 3, 7, 11 and 231
What is the number less than 75 that divides exactly by 7 but leaves remainder 1 when divided by 11?
The numbers that divide evenly by 7 and less than 75 are 7, 14, 21, 28, 35, 42, 49, 56, 63, and 70 Of the above numbers 56 when divided by 11 gives a quotient of 5 and a remainder of 1 Hence the number is 56
Find the greatest number which divides 6168 2447 and 3118 leaving the same remainder in each case?
Well, honey, the greatest number that fits the bill is the difference between the numbers. So, 6168 - 2447 = 3721, and 3118 - 2447 = 671. The greatest number that divides all three and leaves the same remainder is the greatest common divisor of 3721 and 671, which is 671.
What is the number when divided by 5 leaves a remainder of 2 and when divided by 7 leaves a remainder of 5?
3
What does odd number means?
It is an integer which, when divided by 2, leaves a remainder of 1.
What is the remainder of 26 divided by 5?
5.2
What number is between 12 and 32 when divided by 5 it leaves a remainder of 1 when divided by 4 it leaves a remainder of 2?
26
What is the least no when divided by 7 and 8 and 9 respectively leaves remainder 2 and 4and 9 respictively?
No answer is possible as any number divided by 9 must either be exactly divisible by 9 or leave a remainder less than 9.
A no when divided by 114 leaves the remainder 21 if same no is divided by 19 the remainder will be?
2
