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Consult a mathematician
If a three-digit number is divisible by a number greater than 9 and greater than 7, the sum of digits of the three-digit number =9n-2(n is a natural number greater than or equal to 1 and less than or equal to 3).

If a three-digit number is divided by 5 and 2, the mantissa of this three-digit number is 2 or 7.

If a three-digit number is divided by 4 and greater than 3, then the mantissa of this three-digit number must be 7.

Then the sum of the first two digits of this three-digit number is 9n-2-7=9(n- 1).

There is no solution when n= 1.

When n=2, 9(n- 1)=9, including 187, 817,277,727,367,637,457,547 and 907.

When n = 3,9 (n-1) =18, this three-digit number is only1,which is 997.

Because the last two digits -3 can be divisible by 4, those that meet the requirements are187,367,547,727,907.

There are five such three digits.

Sorry, I was careless.