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The only way to get an even number is to put the "4" to the right. To get the largest 3-digit number, you must put the largest digit (5) to the left, so that leaves us with 534.

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Wiki User

8y ago
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SkYy

Lvl 1
1y ago
4,1,6
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Jiesper Khent Quezon

Lvl 1
5mo ago
Form the smallest 3 digit odd number using the digits 3 2 and 7
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Rama Hussein

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5mo ago

534

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Wiki User

7y ago

It is 534.

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SkYy

Lvl 4
1y ago

4

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Q: What is the largest three-digit even number that uses the digits 3 4 and 5?
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Continue Learning about Statistics

I am an even number. I have three digits they are all the same. If you multiply me by 4 all the digits in the product are 8.?

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What is the greatest even two digit number. The product of its digits is 72.?

98


How many two digit even numbers can be composed from nine digits 1 2 3....9?

The first digit can be any one of the nine digits. For each of those . . .The second digit can be any one of four digits . . . 2, 4, 6, or 8 .Total number of possibilities = (9 x 4) = 36


How many even four digit numbers can be made using 02356 where the first digit is cannot be 0 and no digits are repeated?

Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.


How many 4 digit even numbers are there if numbers can be repeated?

Assuming that the second instance of "numbers" in the question means "digits", the answer can be derived as follows: For the number to be even, its last digit must be 0, 2, 4, 6, or 8, for a total of 5 choices. The remainder of the answer depends on whether numbers that begin with 0, 00, or 000 are allowed. If such initial zeroes are not allowed, there are 9 choices for the first digit. Any of the ten digits may be used for the two remaining digits. Therefore the number of instances is 9 X 102 X 5 = 4500. If initial zeroes are allowed, the number of instances is 5000.