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Singapore Math Olympiad

Problem on Prime Numbers | SMO, 2012 | Problem 20

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on Prime numbers.

Problem on Prime Numbers – (SMO Test)


Let A be a 4 – digit integer. When both the first digit (leftmost) and the third digit are increased by n, and the second digit and the fourth digit are decreased by n, the new number is n times A. Find the value of A.

  • 1201
  • 1551
  • 1818
  • 2000

Key Concepts


Algebra

Prime Number

Check the Answer


Answer: 1818

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


If you got stuck you can follow this hint:

We can assume the 4 digit number to be A = \(\overline {abcd}\)

If we expand it into the equation

1000(a+n) + 100(b – n) + 10(c+n) + (d-n) = nA

Try the rest of the sum ………..

After the previous hint :

If we compare the equation it gives :

A + 909 n = nA or

(n-1)A = 909 n

Now one thing we can understand that n and (n-1) are relatively prime and 101 is a prime number . So n= 2 or n= 4.

We have almost got the answer .So try to do the rest now ……….

If n = 4 then A = 1212, which is impossible right?

as b<n given .so

n=2 and A = \( 909 \times 2\) = 1818

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AMC 10

Mean-median – Statistics – AMC 10B, 2019 Problem 13

Mean-median of some numbers


Mean, median, and mode are three kinds of “averages”. … The “mean” is the “average” you’re used to, where you add up all the numbers and then divide by the number of numbers. The “median” is the “middle” value in the list of numbers, And the mode is the number repeated most number of times in the given list. Let’s see how to find the mean-median of some numbers.

Try the problem


What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?

$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$

AMC 10B, 2019 Problem 13

Statistics (Mean-median)

6 out of 10

challenges and thrills of pre college mathematics

Knowledge Graph


mean-median - knowledge graph

Use some hints


The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$.

Now there are only three possibilities for the median, It can be either 6,8 or x. It is because 4 is the smallest number and 17 cannot fit in the middle for any possible value of x.

Now if we consider 6 to be median then we must have to get 6 as the mean also. And we will verify this condition for each of the 6,8, and x.

See the final step for more hints.

So lets start with 6 and then 8 and x itelf.

$\frac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$, so this is a valid solution.

Now let the median be $8$.

$\frac{35+x}{5}=8$ gives $x=5$, so the sequence is $4, 5, 6, 8, 17$, which has median $6$, so this is not valid.

Finally we let the median be $x$.

Finally we let the median be $x$.

$\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$

and the sequence is $4, 6, 8, 8.75, 17$, which has median $8$. This case is therefore again not valid.

Hence the only possible value of $x$ is \((A) -5\).

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AMC 10

Decimal system conversion AMC 10B, 2019 Problem 12

Octal to Decimal system conversion


This problem is based on decimal system conversion. In the given problem we have to convert the given number to the another number system from decimal and finding the sum of the digits later on.

Try the problem


What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$?

$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$

AMC 10B, 2019 Problem 12

Decimal system conversion

6 out of 10

challenges and thrills of pre college mathematics

Knowledge Graph


decimal system conversion- knowledge graph

Use some hints


First of all 2019 is in base 10, see below

\(2019=2*10^3+0*10^2+1*10^1+9+10^0\).

so we have to convert it in the system having base 7.

After converting we will get

$2019_{10} = 5613_7$

But we have to maximize the sum of digits. So we need to increase the number of 6 in the converted number(in base 7) and it is because 6 is the largest number in the number system having base 7.

The conversion of maximize 6 in the number will occur with either of the numbers $4666_7$ or $5566_7$.

and now we can simply find the sum of the digits.

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