Let \(x\) be the least number which, when divided by 16, 24, 30, 36 and 45, leaves a remainder of 4 in each case, and which is divisible by 28. If the HCF of \(x\) and 3193 is \(y\), what is the sum of the digits of \(y\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
A number leaving remainder 4 with each divisor has the form LCM \(\times k+4\); trying \(k=1,2,3,\ldots\) gives the first one divisible by 28. The HCF with 3193 then follows by the division method.
LCM of 16, 24, 30, 36 and 45 \(=2^{4}\times3^{2}\times5=720\)
\(720k+4\) for \(k=1,2,3\): 724, 1444, 2164, none divisible by 28
\(k=4\): \(2884=28\times103\), so \(x=2884\)
\(3193=2884\times1+309\)
\(2884=309\times9+103\)
\(309=103\times3\), so \(y=103\)
Sum of digits of 103 \(=1+0+3=4\)
The sum of the digits of \(y\) is 4.