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


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

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Hundred Integers | ISI-B.Stat Entrance | TOMATO 82

Try this beautiful hundred integers problem based on Remainder useful for ISI B.Stat Entrance.

Hundred Integers | ISI B.Stat Entrance | Problem-82


Let \(x_1,x_2,......,x_100\) be hundred integers such that the sum of any five of them is 20. Then..

  • the largest \(x_i\) equals 5
  • the smallest \(x_i\) equals to 3
  • \(x_{17} = x_{83}\)
  • none of the foregoing statements is true

Key Concepts


Number theory

Divisor

integer

Check the Answer


Answer:\(x_{17} = x_{83}\)

TOMATO, Problem 82

Challenges and Thrills in Pre College Mathematics

Try with Hints


Let us take the numbers be \(x_i , x_j , x_k ,x_l , x_m \)

Now \(x_i + x_j + x_k + x_l + x_m = 20\) and again \(x_i + x_j + x_k + x_l + x_n = 20\)

Can you now finish the problem ..........

From the above relation there are three case arise that....

1)\(x_m = x_n\)

2)All the integers are equal.

3)\(x_{17} =x_{83}\)

So the correct answer is \(x_{17} =x_{83}\)

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Integer Problem | ISI BStat | 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

Check the Answer


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

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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

adition

Cube

Check the Answer


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 \(\frac{108}{6}\)=\(18\)

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