# Find the greatest number which divides 6168, 2447 and 3118 leaving the same reminder in each case.

justaguide | College Teacher | (Level 2) Distinguished Educator

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The greatest number that divides 6168, 2447 and 3118 and leaves the same remainder has to be determined.

Let the number be represented by X. Let the remainder that the number leaves be R.

k1*X + R = 2447

k2*X + R = 3118

k3*X + R = 6168

(k2*X + R) - (k1*X + R) = 3118 - 2447 = 671 = (k1 - k2)*X

The prime factorization of 671 = 11*61. (k2 - k1) and X are whole numbers. For X to take on the largest value, k2 - k1 = 11 and X = 61.

The largest number that leaves the same remainder when 6168, 2447 and 3118 are divided by it is 61.

mathewww | High School Teacher | (Level 1) Honors

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Here, the remainder in each case would be same. So,

6168 - 2447 =3721

6168 - 3118 =3050

3118 - 2447 =671

Now, we have to find the g.c.f. of the numbers 3721,3050 and 671.

g.c.f. of 3721,3050 and 671 = 61

Hence, 61 is the final answer.