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Find the greatest number that will divide 445, 57 and 669 leaving remainders 4, 5, 6 respectively.
$
  (a){\text{ 64}} \\
  (b){\text{ 63}} \\
  (c){\text{ 44}} \\
  (d){\text{ 21}} \\
$


seo-qna
Last updated date: 28th Mar 2024
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MVSAT 2024
Answer
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Hint – In this question use the concept that in order to obtain the greatest number which divides any number (a, b, c) and leaves remainder (x, y, z) should be the highest common factor of the number (a-x, b-y, c-z).

Complete step-by-step solution -
As we know that the numbers say (a, b and c) are exactly divisible by H.C.F of the numbers which is the greatest of the numbers.
Now if we want any remainder suppose (x, y and z) then first we have to subtract (x, y and z) from the given numbers respectively then we calculate the H.C.F of the numbers which is the required largest number.
So the given numbers are (445, 572 and 699).
And we want the remainders (4, 5 and 6).
So first subtract the remainder therefore, we have,
First number = (445 – 4) = 441
Second number = (572 – 5) = 567
Third number = (699 – 6) = 693.
Now calculate the H.C.F of these numbers so first factorize them we have,
Factors of 441 are
$ \Rightarrow 441 = 3 \times 3 \times 7 \times 7$
Factors of 567 are
$ \Rightarrow 567 = 3 \times 3 \times 3 \times 3 \times 7$
Factors of 693 are
$ \Rightarrow 693 = 3 \times 3 \times 7 \times 11$
Now H.C.F of the numbers are the common factors of the numbers.
So the common factors are $3 \times 3 \times 7 = 63$
So required H.C.F is 63.
Thus 63 is the largest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Hence option (B) is correct.

Note – Highest common factor is the largest or the greatest common factor to any two or more given natural numbers. It is also termed as Greatest common divisor that is (G.C.D) so we need to be confused between them.