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

Try this beautiful problem from Number system based on digit problem

## 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 $S$?

- \(31\)
- \(37\)
- \(43\)

**Key Concepts**

Number system

adition

multiplication

## Check the Answer

But try the problem first...

Answer: \(37\)

AMC-10A (2007) Problem 22

Pre College Mathematics

## Try with Hints

First hint

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........

Second Hint

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........

Final Step

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

## Other useful links

- https://www.cheenta.com/problem-based-on-triangle-prmo-2016-problem-10/
- https://www.youtube.com/watch?v=pVg3MHSyt08