+0

Math

0
750
3
the teacher asked ace beatrice and cecil to write down some 4 digit number, write down a second 4 digit number obtained from the first by reversing the order of the digits and then add the two number. aces answer was 5985, beatrice 2212 and cecil 4983.

without looking at their work, the teacher said that both ace and beatrice had mistakes, how could she tell?

if cecil had not made mistakes, what was his initial number, given that it was not divisible by 10 and was greater than the number obtained by digit reversal?
Feb 13, 2012

#2
+5

2991

\$\${\mathtt{2\,991}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,992}} = {\mathtt{4\,983}}\$\$

Start by letting the number be 1000a+100b+10c+d so its reverse is 1000d+100c+10b+a.

Adding these together we get 1001(a+d)+110(b+c).  This must equal 4983.

I'll leave you to take the reasoning further!

May 18, 2014

#1
0

I don't know

May 18, 2014
#2
+5

2991

\$\${\mathtt{2\,991}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,992}} = {\mathtt{4\,983}}\$\$

Start by letting the number be 1000a+100b+10c+d so its reverse is 1000d+100c+10b+a.

Adding these together we get 1001(a+d)+110(b+c).  This must equal 4983.

I'll leave you to take the reasoning further!

Alan May 18, 2014
#3
0

This is another old post!

Never mind, I am going to add the address in with the puzzles.

May 19, 2014