9.......ike 2222 2+2=4 2+2=4
There are 900 of them.
1348.
To form three-digit even numbers from the set {2, 3, 5, 6, 7}, we can use the digits 2 or 6 as the last digit (to ensure the number is even). For each case, we can choose the first two digits from the remaining four digits. For three-digit numbers, there are 2 options for the last digit and (4 \times 3 = 12) combinations for the first two digits, resulting in (2 \times 12 = 24) even numbers. For four-digit even numbers, we again have 2 options for the last digit. The first three digits can be selected from the remaining four digits, giving us (4 \times 3 \times 2 = 24) combinations for each last digit. Thus, there are (2 \times 24 = 48) even four-digit numbers. In total, there are (24 + 48 = 72) three-digit and four-digit even numbers that can be formed from the set.
2
There are 320 such numbers.
There are 900 of them.
1348.
To form three-digit even numbers from the set {2, 3, 5, 6, 7}, we can use the digits 2 or 6 as the last digit (to ensure the number is even). For each case, we can choose the first two digits from the remaining four digits. For three-digit numbers, there are 2 options for the last digit and (4 \times 3 = 12) combinations for the first two digits, resulting in (2 \times 12 = 24) even numbers. For four-digit even numbers, we again have 2 options for the last digit. The first three digits can be selected from the remaining four digits, giving us (4 \times 3 \times 2 = 24) combinations for each last digit. Thus, there are (2 \times 24 = 48) even four-digit numbers. In total, there are (24 + 48 = 72) three-digit and four-digit even numbers that can be formed from the set.
To find the last but one digit in the product of the first 75 even natural numbers, we need to consider the units digit of each number. Since we are multiplying even numbers, the product will end in 0. Therefore, the last but one digit (tens digit) will depend on the multiplication of the tens digits of the numbers. The tens digit will be determined by the pattern of the tens digits of the even numbers being multiplied.
Commonly there are 16 numbers in a credit card. The first digit is the major industry identifier and the first 6 digits including the major industry identifier are the issuer identifier numbers and the last digit is the check digit.
2
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
There are 320 such numbers.
the first number is usually in feet, the last number is in inches. so the last number = 1/12 of the first numbers value.
To find the last digit of ( 21999 \times 32000 ), we can focus on the last digits of the numbers involved. The last digit of ( 21999 ) is ( 9 ) and the last digit of ( 32000 ) is ( 0 ). Multiplying these last digits gives ( 9 \times 0 = 0 ). Therefore, the last digit of ( 21999 \times 32000 ) is ( 0 ).
To form a three-digit even number with different digits using the digits 0, 1, and 2, the last digit must be even, which can only be 0 or 2. If the last digit is 0, the first digit can be either 1 or 2 (2 options), and the middle digit will take the remaining digit (1 option). This gives us 2 valid numbers: 120 and 210. If the last digit is 2, the first digit can only be 1 (since it cannot be 0), and the middle digit must be 0. This gives us 1 valid number: 102. In total, there are 3 different three-digit even numbers: 120, 210, and 102.