The largest three-digit number with a digit sum of 10 is 910. This can be determined by maximizing the hundreds place first, setting it to 9, and then adjusting the tens and units places to reach a total digit sum of 10. In this case, using 1 for the tens place and 0 for the units place achieves the desired sum.
105 is the sum
24
Add the last digit (units digit) to twice the previous digit (tens digit). If this sum is divisible by 4, so is the original number.
Seven of them.
The units digit of a two digit number exceeds twice the tens digit by 1. Find the number if the sum of its digits is 10.
The largest three-digit number with a digit sum of 10 is 910. This can be determined by maximizing the hundreds place first, setting it to 9, and then adjusting the tens and units places to reach a total digit sum of 10. In this case, using 1 for the tens place and 0 for the units place achieves the desired sum.
105 is the sum
105 is the sum
"If the units digit and the hundreds digit of the number 513 were reversed..." 315 'find the sum of the original number and the new number." 513+315=828
-4
24
Add the last digit (units digit) to twice the previous digit (tens digit). If this sum is divisible by 4, so is the original number.
When the units digit equals the tens digit then the sum of the digits of a 2 digit number is double the units digit. In each tens range above 50, numbers below this critical point meet the requirement, numbers above this critical point have a sum LESS than double the units digit. The applicable numbers are 51-54 (4), 61-65 (5), 71-76 (6), 81-87 (7) and 91-98 (8). Then there are 4 + 5 + 6 + 7 + 8 = 30 qualifying numbers.
31
Seven of them.
46