Categories

## 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

multiplication

Answer: $37$

AMC-10A (2007) Problem 22

Pre College Mathematics

## Try with Hints

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

Categories

## Numbers on cube | AMC-10A, 2007 | Problem 11

Try this beautiful problem from AMC 10A, 2007 based on Numbers on cube.

## Numbers on cube – AMC-10A, 2007- Problem 11

The numbers from $1$ to $8$ are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

• $16$
• $18$
• $20$

### Key Concepts

Number system

Cube

Answer: $18$

AMC-10A (2007) Problem 11

Pre College Mathematics

## Try with Hints

Given condition is “The numbers from $1$ to $8$ are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same”.so we may say that if we think there is a number on the vertex then it will be counted in different faces also.

can you finish the problem……..

Therefore we have to count the numbers $3$ times so the total sum will be $3(1+2+….+8)$=$108$

can you finish the problem……..

Now there are $6$ faces in a Cube…..so the common sum will be $\frac{108}{6}$=$18$

Categories

## Integer Problem |ISI-B.stat | Objective Problem 156

Try this beautiful problem Based on Integer, useful for ISI B.Stat Entrance.

## Integer | ISI B.Stat Entrance | Problem 156

Let n be any integer. Then $n(n + 1)(2n + 1)$

• (a) is a perfect square
• (b) is an odd number
• (c) is an integral multiple of 6
• (d) does not necessarily have any foregoing properties.

### Key Concepts

Integer

Perfect square numbers

Odd number

Answer: (c) is an integral multiple of 6

TOMATO, Problem 156

Challenges and Thrills in Pre College Mathematics

## Try with Hints

$n(n + 1)$ is divisible by $2$ as they are consecutive integers.

If $n\not\equiv 0$ (mod 3) then there arise two casess……..
Case 1,,

Let $n \equiv 1$ (mod 3)
Then $2n + 1$ is divisible by 3.

Let $n \equiv2$ (mod 3)
Then$n + 1$ is divisible by $3$

Can you now finish the problem ……….

Now, if $n$ is divisible by $3$, then we can say that $n(n + 1)(2n + 1)$ is always
divisible by $2*3 = 6$

Therefore option (c) is the correct