Daniel's integer reversal
How many 3-digit positive integers N are there such that if M is the integer obtained by reversing the digits of N , then
M = 3 8 2 7 N ?
This problem is posed by Daniel W .
The answer is 2.
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4 solutions
I thought, Let N = a b c so M = c b a we get, N − M = 9 9 ( a − c ) ⟹ N − ( 2 7 / 3 8 ) N = 9 9 ( a − c ) ⟹ N = 9 ⋅ 3 8 ⋅ ( a − c ) As, N is 3 digit, ( a − c ) can be 1 or 2 that gives N = 9 ⋅ 3 8 ⋅ 1 = 3 4 2 and 9 ⋅ 3 8 ⋅ 2 = 6 8 4 No guessing needed ;)
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Nice solution!
Great!
I don't really understand what the symbol stands for.
Such a simple yet thoughtful solution. Great job..
danielIntegerReversal.c
#include <stdio.h>
int reverse(int n) {
int result = 0;
while (n > 0) {
result = result * 10 + n%10;
n = n/10;
}
return result;
}
int main(char* args, int nArgs) {
int n,results = 0;
float m,condition = 0;
for (n=100; n < 1000; n++) {
m = (float) reverse(n);
condition = n*(27/38.0);
if (condition == m) {
printf("m=%f %d*(27/28)=%f !\n",
m,
n,
condition);
results++;
}
}
printf("found %d numbers.\n",results);
}
m=243.000000 342*(27/28)=243.000000 !
m=486.000000 684*(27/28)=486.000000 !
found 2 numbers.
In python:
numbs = [i for i in range(100,1000)]
for i in range(0,900):
x = str(numbs[i])
new_num = ""
new_num = new_num + x[2]
new_num = new_num + x[1]
new_num = new_num + x[0]
if 38*int(new_num) == 27*numbs[i]:
count = count + 1
print(count)
which prints 2...
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ha, custom made reverse using str() and int(). not very elegant, but works.
from the text of the question you will figure out that "m/n=27/38" ===> 100x+10y+z/100z+10y+x=27/38 ( Without any Problem! :-) ) so: 2700z+270y+27x=3800x+380y+38 ====> 3773x+110y=2662z ====><devide by 11> 343x+10y=242z and then you can guess the numbers(That I think is not very good idea! ) or continue...
Finally The nums whould be 342 & 684. >>>>2 is the answer.
M is an integer so N is a multiple of 3 8 . M is a multiple of 2 7 and hence N is a multiple of 9 because the digit sum will be the same.
Clearly N is a multiple of 2 , 9 , and 1 9 so N is either 3 4 2 or 6 8 4 and it can be quickly verified that both of these values of N will generate values of M that follow the rules as given, so the answer is 2
2 7 ∣ M ⟹ 9 ∣ M ⟹ 9 ∣ S ( M ) = S ( N ) , where S ( x ) denotes the digit sum of x . Also, 9 ∣ S ( N ) ⟹ 9 ∣ N ⟹ 9 ⋅ 3 8 = 3 4 2 ∣ N , as 3 8 ∣ N . We obtain that N ∈ { 3 4 2 , 6 8 4 } ⟹ ( M , N ) ∈ { ( 2 4 3 , 3 4 2 ) , ( 4 8 6 , 6 8 4 ) } , of which both pairs satisfy the constraints.