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.

Related questions

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.

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

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

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

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

The first digit can be any one of 9 digits (anything but zero). For each of those ...The second digit can be any one of 9 digits (anything but the previous one). For each of those ...The third digit can be any one of 9 digits (anything but the previous one). For each of those ...The fourth digit can be any one of 9 digits (anything but the previous one). For each of those ...The fifth digit can be any one of 9 digits (anything but the previous one).So the total number of possible arrangements is (9 x 9 x 9 x 9 x 9) = 59,049out of a total of 99,999 different 5-digit numbers, or 59.05%.

1

There are 90 palindromes with 4 digits.The first digit can be any digit from the set {1,2,3,4,5,6,7,8,9}.With each choice of the first digit, the second can be any digit from the set {0,1,2,3,4,5,6,7,8,9}.That makes 9*10 = 90 permutations for the first two digits. These determine the palindrome since the third and fourth digits are the same as the second and first, respectively.