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.
6
Find the greatest number which divides 742 and 1162find the greatest number which divides 742 and 1162 117
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
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
5.2
26
No answer is possible as any number divided by 9 must either be exactly divisible by 9 or leave a remainder less than 9.
56
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.
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
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
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
3
It is an integer which, when divided by 2, leaves a remainder of 1.
5.2
26
No answer is possible as any number divided by 9 must either be exactly divisible by 9 or leave a remainder less than 9.
26. This is because when you divide 26 by 5, you have a remainder of 1, and when you divide it by 4, you have a remainder of 2
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.