Just count up ur 0's
and put it up like this example:
we have the number
5000 wich has three 0's. We then just make a normal out of it wich in this case is:
5*10^3 or if u prefer 5E+3
There is no four digit number where the ones is twice the tens, the hundreds is five less than the ones, and the thousands is the sum of the tens and hundreds. int ones, tens, hundreds, thousands; for (thousands=1; thousands<10; thousands++) { /**/ for (hundreds=0; hundreds<10; hundreds++) { /**/ /**/ for (tens=0; tens<10; tens++) { /**/ /**/ /**/ for (ones=0; ones<10; ones++) { /**/ /**/ /**/ /**/ if (ones != 2 * tens) break; /**/ /**/ /**/ /**/ if (hundreds != ones - 5) break; /**/ /**/ /**/ /**/ if (thousands != tens + hundreds) break; /**/ /**/ /**/ /**/ printf ("dd\n", thousands, hundreds, tens, ones); /**/ /**/ /**/ } /**/ /**/ } /**/ } }
It is 9326.
13,432
50 thousands, 3 hundreds, 7 tens, 2 ones
I don't believe this is possible.
There is no four digit number where the ones is twice the tens, the hundreds is five less than the ones, and the thousands is the sum of the tens and hundreds. int ones, tens, hundreds, thousands; for (thousands=1; thousands<10; thousands++) { /**/ for (hundreds=0; hundreds<10; hundreds++) { /**/ /**/ for (tens=0; tens<10; tens++) { /**/ /**/ /**/ for (ones=0; ones<10; ones++) { /**/ /**/ /**/ /**/ if (ones != 2 * tens) break; /**/ /**/ /**/ /**/ if (hundreds != ones - 5) break; /**/ /**/ /**/ /**/ if (thousands != tens + hundreds) break; /**/ /**/ /**/ /**/ printf ("dd\n", thousands, hundreds, tens, ones); /**/ /**/ /**/ } /**/ /**/ } /**/ } }
7,478 is how you right 7 thousands, 4 hundreds, 6 tens, 18 ones.
It is 9326.
13,432
7 ten thousands 9 hundredths 3 ones 6 hundreds 7 tens = 71,573
608,907
7252
tens come after ones then hundreds then thousands
80,400,295
50 thousands, 3 hundreds, 7 tens, 2 ones
I don't believe this is possible.
40,683 ********************** I think it should be: 46,083 (There are no hundreds.)