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Q: What is a two digit number that is one third of the number is 7 more than one fourth of the number and the first digit is twice the second digit and the sum of the digits is a two digit number?

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There are a lot of possibilities. The second digit can be 2 through 6, the third digit can be 3 through 7 as long as it is larger than the second digit. What we have so far: 1 _ _ 89

There are 400. Assuming the number must be at least 10,000, then: In a 5 digit palindrome, the first and last digits must be the same, and the second and fourth digits must be the same; and: For the first and last digit there is a choice of 4 digits {2, 4, 6, 8}; For each of these there is a choice of 10 digits {0, 1, ..., 9} for the second and fourth digits; For each of the above choices these is a choice of 10 digits {0, 1, ..., 9} for the third digit; Making 4 x 10 x 10 = 400 possible even 5 digit palindromes.

The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5

Just compare the digits one by one: compare the first digit after the decimal point with the first digit of the other number, the second digit with the second digit, etc., until you find a digit that is different.

A number is made up from digits in the numeral system. We often use the decimal system in which we use 10 digits, In writing the any number, many digits are used, even repitation of digits. when we write any number using the digits, the last digit ( from right side ) in that number is called unit digit. for example in the number 9814868980, here 0 is called unit digit.

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Not necessarily. Consider 444. The digits are not different. The first and second digits are not multiples of 3 The first digit is not greater than the second digit. In spite of all that, 444 is a 3-digit number

There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.

The first and last, and the second and third.

There are a lot of possibilities. The second digit can be 2 through 6, the third digit can be 3 through 7 as long as it is larger than the second digit. What we have so far: 1 _ _ 89

There are 400. Assuming the number must be at least 10,000, then: In a 5 digit palindrome, the first and last digits must be the same, and the second and fourth digits must be the same; and: For the first and last digit there is a choice of 4 digits {2, 4, 6, 8}; For each of these there is a choice of 10 digits {0, 1, ..., 9} for the second and fourth digits; For each of the above choices these is a choice of 10 digits {0, 1, ..., 9} for the third digit; Making 4 x 10 x 10 = 400 possible even 5 digit palindromes.

27

To confirm the validity of the DEA registration number, the pharmacist should add the first, third, and fifth digits together, then add the second, fourth, and sixth digits, multiplying that sum by 2. The right-most digit of the sum of these 2 calculations will correspond with the final, or seventh, digit of a valid number.

The first digit can be any one of 10 digits. For each of those . . .The second digit can be any one of the remaining 9 digits. For each of those . . .The third digit can be any one of the remaining 8 digits. For each of those . . .The fourth digit can be any one of the remaining 7 digits.The total number of possibilities is (10 x 9 x 8 x 7) = 5,040

1155

1349

1

First digit can be any of 3, second can be any of 3, as can third and fourth so your answer is 3 to the fourth ie 81