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Q: The units digit of a two digit number exceeds twice the tens digit by 1 what numbers if the sum of its digit is 10?

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If the number is odd, the last digit is odd. That means the ten's digit must be twice the one's digit. There are only four two-digit numbers where the ten's digit is twice the one's digit: 21, 42, 63, 84. Check which of these satisfy all the clues.

Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.

18 and 38

18 and 38

None. The sum of one digit can't be twice the size of the digit.

120 5-digit numbers can be made with the numbers 12345.

A proper 2-digit number (which must be smaller than 100) cannot be twice a 10-digit number (which must be larger than 999,999,999).

216.

216

Well 18 works. [1+8 = 9]. It's not exactly the only one, though. Consider zero. The sum of the digits (0) is zero. And twice zero = zero.Here are some ideas to see if there are others: For 1-digit numbers, the number is the sum, so the zero, above, is the only one that satisfies the conditions. For 2-digit numbers, the largest sum possible is 18 [99: 9 + 9]. Twice that sum is 36, so no numbers greater than 36 (with 2-digit numbers).Now consider the unknown number with digits AB. So the sum = A + B, and the number = 10*A + B. Now the number is twice the sum, so:10*A + B = 2*(A + B). Rearranging, we have 8*A = B, so you can substitute whole numbers for A, and calculate B. There is the trivial case of zero: 8*0 = 0, which was already covered. If A = 1, then B = 8, which gives us 18. With A = 2, then B= 16, so that's not a 2-digit number. So 18 is the only 2-digit number.What about 3-digits: the largest 3-digit sum is 9+9+9=27. Twice that is 54, which is a 2-digit number. So there can be no 3-digit numbers, satisfying the conditions.

The largest six digit number period is 999999. The largest number meeting your criteria is 998949.

1

100012

There are 360 of them.

2.3

17

In order to get the greatest number we want to try and get the biggest possible numbers in the hundred thousands, ten thousands and thousand places. There is no rule against the hundred thousands digit being 9 so it must be 9. The ten thousand digit needs to be twice the tens so it must be 8 while the tens is 4. The thousands digit is divided by the units, and the number is even. So thousands must be 6 and the units must be 2. Thus the hundreds must be 7. The number is 986,742

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.

AB If B = x, then A = x + 4 Also, A = 2x -1 So that, 2x - 1 = x + 4 x = 5 B = 5, A = 9 So AB = 95

The number that matches your criteria is... 998949

22621, 44742, 66863, 88984

100,000

4284

102, 010 2 in the thousands place is twice 1 in the tens place,

11,33

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