#### You may also like ### Counting Factors

Is there an efficient way to work out how many factors a large number has? ### Summing Consecutive Numbers

15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers? ### Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

# Arrange the Digits

##### Age 11 to 14 Challenge Level:

The closest total is 1503.

The first two correct answers to this problem were received from Natalie and Jake, both from the West Flegg GM Middle School. Both had arrived at the answer by trial and improvement techniques.

Later on, work by Thomas from Wymondham High School arrived. He had been a little more systematic in his search and explained how he strove initially to get the hundreds column to total 15 and the tens column to total 9. Unsuccessful at this he moved on to consider making the units column at least 20 and the tens column eight, which after a little searching gave him a result that he was satisfied with.

In whatever way the digits 1 to 9 are arranged and added together the total will always have a digital root that is 9.

e.g. 123 + 45 + 678 + 9 = 855

where 8 + 5 + 5 = 18, i.e the sum of the digits

and 1 + 8 = 9

123 + 456 + 789 = 1368

and the sum of the digits is 1 + 3 + 6 + 8 = 18

and 1 + 8 = 9

When restricted to three 3-digit numbers, they too must have a total whose digital root is 9.

Hence, 1500 cannot be a solution because its digital root is 6. While the integers 1494 and 1503 both have digital roots of 9, 1503 is nearer to 1500 than 1494.

Search for the digits which would make the units add up to 13 or 23, the tens digits add up to 9 or 8 and the hundreds digits add up to 13 or 14 e.g. 519 + 748 +236 = 1503.

See the article Divisibility Tests.