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JEE Advanced 2019 · Paper 1, Mathematics question 14

Let AP(a;d)AP(a; d) denote the set of all the terms of an infinite arithmetic progression with first term aa and common difference d>0d > 0. If

AP(1;3)∩AP(2;5)∩AP(3;7)=AP(a;d)AP(1;3)\cap AP(2;5)\cap AP(3;7) = AP(a;d)

then a+da + d equals ______

Answer and worked solution

Official answer: 157

We need x≡1(mod3)x\equiv1\pmod3, x≡2(mod5)x\equiv2\pmod5 and x≡3(mod7)x\equiv3\pmod7.

Solve. The numbers ≡3(mod7)\equiv3\pmod7 that are ≡2(mod5)\equiv2\pmod5 are 17,52,87,…17,52,87,\dots. Among these, 52≡1(mod3)52\equiv1\pmod3. So x≡52(mod105)x\equiv52\pmod{105}.

The intersection. It is AP(52;105)AP(52;105), since 52 already exceeds every first term. So a+d=52+105=157a+d=52+105=157.

Answer: 157