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Digits of number | PRMO 2018 | Question 3

Try this beautiful problem from the PRMO, 2018 based on Digits of number.

Digits of number - PRMO 2018


Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.

  • is 107
  • is 70
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Numbers

Multiples

Check the Answer


Answer: is 70.

PRMO, 2018, Question 3

Higher Algebra by Hall and Knight

Try with Hints


abccba (b is odd)

=a(10^5+1)+b(10^4+10)+c(10^3+10^2)

=a(1001-1)100+a+10b(1001)+(100)(11)c

=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c

=7p+(c-a) where p is an integer

Now if c-a is a multiple of 7

c-a=7,0,-7

hence number of ordered pairs of (a,c) is 14

since b is odd

number of such number=14 \times 5=70.

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Try this beautiful problem from the PRMO, 2018 based on Digits of number.

Digits of number - PRMO 2018


Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.

  • is 107
  • is 70
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Numbers

Multiples

Check the Answer


Answer: is 70.

PRMO, 2018, Question 3

Higher Algebra by Hall and Knight

Try with Hints


abccba (b is odd)

=a(10^5+1)+b(10^4+10)+c(10^3+10^2)

=a(1001-1)100+a+10b(1001)+(100)(11)c

=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c

=7p+(c-a) where p is an integer

Now if c-a is a multiple of 7

c-a=7,0,-7

hence number of ordered pairs of (a,c) is 14

since b is odd

number of such number=14 \times 5=70.

Subscribe to Cheenta at Youtube


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