If you use the word "and" in naming numbers above 100, then, for example, 999 is named "nine hundred and ninety-nine". Considering this as four words, one of them hyphenated, the letter counts in the four words are 4, 7, 3 and 10. The product of these four numbers is 840.
There is one 3-digit number such that if you apply the above procedure, the final product of its four word lengths is the same as the original number. What is that number?
sol:
p=A*B*C
A is number of hundreds , 3,4,or 5 letters
B is 21=7*3 (product of # of letters in "hundred and")
C is # of letters in the two last words (8,9,10 and 11)
There are at most 15 possible results to be checked from 3*21*8 to 5*21*11.
3*21*8=504 4*21*5=420
3*21*9=567 4*21*10=840
3*21*10=630 3*21*6=378
3*21*11=693 3*21*11=693 answer STOP
trial end error
