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Ans to Q1. The approach is, let the number be N. We have

N=d1q1+r1 and N=d2q2+r2. d1 and d2 are divisors. q1 & q2 are quotients and r1&7 r2 are remainders. Its given that

N=4q1+1 and N=7q2+4. Where q1 and q2 are quotients. 4 and 7 are divisors and 1 and 4 are remainders.

Substituting q1=1,2,3,4,5, and 6, we have, N=5 or 9 or 13 or 17 or 21 or 25
substituting q2=1,2,3, we have N=11 or 18 or 25. So N=25 is common to both divisors 4 and 7. So the answer is 25.

For Q2: We can solve in two ways. Method 1: Apply one of the popular rules.
The rule says, if we have three divisors (d1, d2 and d3) and three remainders (r1, r2 and r3), then the complete remainder is given by the formula d1d2r3 + d1r2 + r1.

Note that 84 is a product of 3, 4 and 7.
Here we have d1=3, d2=4, d3=7 and r1=2, r2=1 and r3=4. Applying the rule we have
48 + 3 + 2 = 53. Hence the answer is 53

Alternate Method 2: (easier one - need not remember any formula and is a technique based approach. Dont ask me to prove it. But works all the time)

We know N= 3q1 + 2 or 4q2 + 1 or 7q3 + 4.

let q3=0, we then have 7q3+4 = 4.
substitute 4 as q2. We then have 4q2 + 1 = 17.
substitute 17 as q1. We then have 3q1 + 2 = 53. So N= 53. 53 when divided by 84 leave remainder 53.

I will make it even better by extrapolating what has been done above.

Let q3=1, we have 7q3+4 = 11
substitute 11 as q2. We then have 4q2 + 1 = 45.
substitute 45 as q1. We then have 3q1 + 2 = 137. So N= 137. 137 when divided by 84 leave remainder 53.

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