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The answer is 33
Lets put them into equations
172 = D*q1 + r1;
205 = D*q2 + r2;
304 = D*q3 + r3;
For the greatest value of D take the HCF of 33, 99 and 132 which is 33.
step1 let assume a=172
step2 take difference
now take a HCF of these numbers(33,99,132)
33 is the answer
The greatest number that will divide 172,205 and 304 leaving the same remainder in each
case is HCF ((205-172), (304-205), (304-172))
= HCF of 33, 99, and 132
99=3 x 3 x 11
132=3 x 2 x 2 x11
Required number =3 x 11=33
A number when divided by 48 gives 47 as remainder and x as quotient, the same number when divided by 49 gives 8 as remainder and y as quotient. What is the minimum possible value of x?
My house number is such that, when divided by 2, 3, 4, 5 or 6 it will always leave a remainder of 1.
However, when divided by 11 there is no remainder.
What is my house number?
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