Number system | AMC-10A, 2007 | Problem 22

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let \(S\) be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$?

The given condition is "A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term,And also another codition that the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term" so we may assume four integers that be \((xyz,yzm,zmp,qxy)\) i.e\((100x+10y+z,100y+10z+m,100z+10m+p,100q+10x+y)\)

Now the sum of the digits be\((110x+111y+111z+11m+p+100q)\)

can you finish the problem........

But "the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term"......so we may say that in last integer \(qxy\)...\(q=m\) & \(p=x\).Therefore the sum becomes \((110x+111y+111z+11q+x+100q)\)=\(111(x+y+z+m)\) i.e \(111 K\) (say)

can you finish the problem........

N ow in \(111K\)= \(3.37.K\).........So in the given answers the largest prime number is 37

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