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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;

and r1=r2=r3=r

so. D*(q2-q1)= 33 D*(q3-q2)= 99 D*(q3-q1)= 132

For the greatest value of D take the HCF of 33, 99 and 132 which is 33.

step1 let assume a=172 b=205 c=304 step2 take difference (b-a)=33 (c-b)=99 (c-a)=132 now take a HCF of these numbers(33,99,132) HCF=33(ans)

33 is the answer Explanation: 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 33=31x11 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?