একটা সংখ্যাকে বিলম্বী-কিশোর বলা হবে যদি সেটা তার অঙ্কগুলোর যোগফলের \(19\) গুণ হয়। কতগুলো বিলম্বী-কিশোর সংখ্যা আছে?
A late-teen number is a positive integer which is $19$ times the sum of its own digits. Determine how many late-teen numbers are there.
BdMO National 2021 Junior Problem 7
- Anindya Biswas
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"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Anindya Biswas
- Posts:264
- Joined:Fri Oct 02, 2020 8:51 pm
- Location:Magura, Bangladesh
- Contact:
Re: BdMO National 2021 Junior Problem 7
Detailed solution :Anindya Biswas wrote: ↑Mon Apr 12, 2021 11:54 amএকটা সংখ্যাকে বিলম্বী-কিশোর বলা হবে যদি সেটা তার অঙ্কগুলোর যোগফলের \(19\) গুণ হয়। কতগুলো বিলম্বী-কিশোর সংখ্যা আছে?
A late-teen number is a positive integer which is $19$ times the sum of its own digits. Determine how many late-teen numbers are there.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
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- Joined:Fri Jan 28, 2022 4:29 pm
Re: BdMO National 2021 Junior Problem 7
Is there any easier method?
Re: BdMO National 2021 Junior Problem 7
Let the 3 digit number be (100a+10b+c) were a,b & c are integers between 0 and 9dinosaur game, and a>0 .
The condition that the number is 19 times greater than sum of its digit is
(100a+10b+c)=19(a+b+c)
or
81a−9b−18c=0
or
9a−b−2c=0
or
a=b+2c9
Now we have to select b and c (between 0 and 9 ) such (b+2c) is divisible with 9
There are finite possibilities.
c=0 , gives b=9 and a=1 (Note: b=0 gives a=0 which violates a>0 condition)
c=1 gives b=7 and a=1
c=2 gives b=5 and a=1
c=3 gives b=3 and a=1
c=4 gives b=1 and a=1
c=5 gives b=8 and a=2
c=6 gives b=6 and a=2
c=7 gives b=4 and a=2
c=8 gives b=2 and a=2
c=9 gives b=0 and a=2 also
c=9 gives b=9 and a=3
So there are 11 possibilities. The numbers are
(1) 190
(2) 171
(3) 152
(4) 133
(5) 114
(6) 285
(7) 266
(8) 247
(9) 228
(10) 209
(11) 399
The condition that the number is 19 times greater than sum of its digit is
(100a+10b+c)=19(a+b+c)
or
81a−9b−18c=0
or
9a−b−2c=0
or
a=b+2c9
Now we have to select b and c (between 0 and 9 ) such (b+2c) is divisible with 9
There are finite possibilities.
c=0 , gives b=9 and a=1 (Note: b=0 gives a=0 which violates a>0 condition)
c=1 gives b=7 and a=1
c=2 gives b=5 and a=1
c=3 gives b=3 and a=1
c=4 gives b=1 and a=1
c=5 gives b=8 and a=2
c=6 gives b=6 and a=2
c=7 gives b=4 and a=2
c=8 gives b=2 and a=2
c=9 gives b=0 and a=2 also
c=9 gives b=9 and a=3
So there are 11 possibilities. The numbers are
(1) 190
(2) 171
(3) 152
(4) 133
(5) 114
(6) 285
(7) 266
(8) 247
(9) 228
(10) 209
(11) 399