For how many Numbers less than 100 is the digit in the ones place half the digit in the tens place?

Answer 1

21, 42, 63, 84

Answer 2

4

Answer 3

The question does not ask about numbers between 0 and 100, it merely asks about numbers less than 100, which means that there are no lower bounds. Nor does it specify integers.

So, generalising the lines along which the previous contributor is thinking ...

It follows that numbers that satisfy the stated conditions (rather than the assumed ones) must necessarily include the following subsets of numbers:

(a) numbers ending in decimals: 42.1, 42.11, and 42.111 etc
(b) numbers ending in fractions eg 42 1/5, 42 1/6, 42 1/7 etc
(c) negative 2-digit numbers eg -42, -63, -84 etc
(d) negative 2-digit numbers with fractions
(e) negative 2-digit numbers with decimals
plus of course
(f) negative numbers of any number of digits greater than 2 eg -55542, -55642, -55742 etc
(g) negative numbers of any number of digits greater than 2 with fractions
(h) negative numbers of any number of digits greater than 2 with decimals

i.e. an infinite number of solutions exists.

Answer 4

The digit in the tens place must be even.

I assume you mean nonzero numbers, 1,2,...,99.

You have 21, 42, 63, 84. So there are 4.

Answer 5

I'm assuming we're talking only whole numbers here...

In that case, there are only 4 such numbers: 84, 63, 42, 21

You certainly could have figured that out quicker than it took you to type the question on the computer. Don't be lazy...impress your teacher with your hard work.

Answer 6

99

Answer 7

total numbers total digits = 10 for 4 digit volume 0 should not be waiting to be utilized in 1000's place you need to use any of (a million to 9) 9 digits in 1000's place you need to use any of very final 9 digits in one hundred's place you need to use any of very final 8 digits in 10's place you need to use any of very final 7 digits in a million's place total numbers shaped = 9 × 9 × 8 × 7 = 4536 For numbers divisible by skill of 5 you could desire to placed 0 or 5 in gadgets place you could desire to no longer have the ability to place 0 or 5 in one thousand' or one hundred or 10's place you need to use any of 8 digits in 1000's place you need to use any of very final 7 digits in one hundred's place you need to use any of very final 6 digits in 10's place you need to use any of very final 2 digits in a million's place total numbers shaped = 8 × 7 × 6 × 2 = 672 -----

Answer 8

84, 63, 42, 21
Four

Answer 9

21
42
63
84

Total 4

Answer 10

There's an infinite amount of them.

Some examples include 42, 42.1, 42.2, -42.1, and -42.2