Chat with our AI personalities
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); /**/ /**/ /**/ } /**/ /**/ } /**/ } }
Subtracting 15 ones from 15 tens involves converting the tens into ones to perform the subtraction. Since each ten is equivalent to 10 ones, subtracting 15 ones from 15 tens means subtracting 150 from 15, resulting in -135. This is because after converting 15 tens to ones, you have a total of 150 ones, and subtracting 15 ones leaves you with 135 ones.
What number has 9 tens and 4 fewer ones than tens
22 tens plus 6 ones.
The general function is:1. y = a*x+bb is irrelevant and we can be removed2. y = a*xlets split x into ones and tens3. x = tens*10 + ones /e.g. 23 = 2*10 + 34. p1 = Multiplier of the onesp2 = Multiplier of the tens5. y = tens*10*p2 + ones*p1 /according to the question6. x*a = tens*10*p2 + ones*p1 /according to 2.7. (tens*10 + ones)*a = tens*10*p2 + ones*p1 /according to 3.8. tens*10*a + ones*a = tens*10*p2 + ones*p1 /regroup9. tens*10*a - tens*10*p2 + ones*a - ones*p1 = 0 /regroup10. tens*10*(a-p2) + ones*(a-p1) = 0 /regroup11. assuming "tens" and "ones" are not 0 then (a-p2) and (a-p1) must be 012. a-p2 = 0a-p1 = 013. a = p2a = p114. a = p1 = p2the answer is: when the Multipliers of ones and tens are equal then the product is called a.