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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.

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Find the greatest number which divides 742 and 1162find the greatest number which divides 742 and 1162 117

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Q: Find the greatest number which divides 742 and 1162 leaves a remainder of exactly 7 in each case?
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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 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 is the remainder of 26 divided by 5?

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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?

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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.

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What is the number less than seventy-five that divides exactly by seven but leaves remainder one when divided by eleven?

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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


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 is the number when divided by 5 leaves a remainder of 2 and when divided by 7 leaves a remainder of 5?

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The number you are thinking of is between 12 and 32 when you divide it by 5 it leaves a remainder of 1 when i divide it by 4 it leaves a remainder of2?

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What is the greatest number which when divided by 2053 and 967 leaves a remainder of 5 and 7 respectively?

The smallest number that satisfies the problem is 1718366. There is no greatest number; whatever value you specify as the greatest I can add 967 × 2053 = 1985251 and get an even greater number that solves the problem. This addition is divisible by both 2053 and 967, so when dividing by 2053 or 967 the (respective) remainder does not change.