Difference between revisions of "2013 AIME I Problems/Problem 2"

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== Solution==
 
== Solution==
 
The number takes a form of <math>5\text{x,y,z}5</math>, in which <math>5|x+y+z</math>. Let <math>x</math> and <math>y</math> be arbitrary digits. For each pair of <math>x,y</math>, there are exactly two values of <math>z</math> that satisfy the condition of <math>5|x+y+z</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math>
 
The number takes a form of <math>5\text{x,y,z}5</math>, in which <math>5|x+y+z</math>. Let <math>x</math> and <math>y</math> be arbitrary digits. For each pair of <math>x,y</math>, there are exactly two values of <math>z</math> that satisfy the condition of <math>5|x+y+z</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math>
 +
 +
This 9-line code in Python also gives the answer too.
 +
import math
 +
counter=0
 +
for integer in range(10000,99999):
 +
  if str(integer)[4] != '5' or str(integer)[0] != '5':
 +
    counter=counter
 +
  else:
 +
    if math.remainder(int(str(integer)[1]+str(integer)[2]+str(integer)[3]),5)==0:
 +
      counter+=1
 +
print(counter)
  
 
==Video Solution==
 
==Video Solution==

Revision as of 01:30, 3 October 2020

Problem 2

Find the number of five-digit positive integers, $n$, that satisfy the following conditions:

    (a) the number $n$ is divisible by $5,$
    (b) the first and last digits of $n$ are equal, and
    (c) the sum of the digits of $n$ is divisible by $5.$


Solution

The number takes a form of $5\text{x,y,z}5$, in which $5|x+y+z$. Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$, there are exactly two values of $z$ that satisfy the condition of $5|x+y+z$. Therefore, the answer is $10\times10\times2=\boxed{200}$

This 9-line code in Python also gives the answer too. import math counter=0 for integer in range(10000,99999):

 if str(integer)[4] != '5' or str(integer)[0] != '5':
   counter=counter
 else:
   if math.remainder(int(str(integer)[1]+str(integer)[2]+str(integer)[3]),5)==0:
     counter+=1

print(counter)

Video Solution

https://www.youtube.com/watch?v=kz3ZX4PT-_0 ~Shreyas S

See also

2013 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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