I am thinking of some #s less than 100. When I divide them by 2 there is a remainder of 1. When I divide them by 3 there is a remainder of 1. When I divide it by 4 there is a remainder of 1. However, when I divide by 5 there is no remainder. what are the #'s?
- my answer so far is every number ending in a zero or five from 1-100 but I don't think its that simple. If someone could help me figure this out it would be greatly appreciated!
The statement, "when I divide them by 2, there is a remainder of 1" means that the numbers must be odd, therefore they can't end in a 0. I'm only seeing 2 answers, 25 and 85
like Joe said, the first condition limits you to odd numbers
the second condition limits you to the numbers
7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 (every odd multiple of 3 + 1)
the third condition limit you to the numbers
5 9 13 1721 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97 ( every multiple of 4 + 1)
the fourth conditon gives you
5 15 25 35 45 55 65 75 85 95
As Joe pointed out , the only numbers fitting all 4 conditions are
25 and 85
To find the numbers that satisfy the given conditions, we need to find numbers that leave a remainder of 1 when divided by 2, 3, and 4, but leave no remainder when divided by 5.
Let's break this down step by step:
1. Start by listing the multiples of 2, 3, and 4 individually, until the pattern repeats or reaches a number close to 100:
Multiples of 2: 2, 4, 6, 8, 10, 12, 14, ...
Multiples of 3: 3, 6, 9, 12, 15, ...
Multiples of 4: 4, 8, 12, 16, ...
2. Notice that every multiple of 12 satisfies the conditions because it leaves a remainder of 1 when divided by 2, 3, and 4.
3. Concerning the condition of leaving no remainder when divided by 5, we know that the last digit of the number must be either 0 or 5. However, not all multiples of 12 satisfy this condition.
4. To find the multiples of 12 that also leave no remainder when divided by 5, we can check each number individually by dividing it by 5 and checking if the remainder is 0:
12 ÷ 5 = 2 remainder 2 (not divisible by 5)
24 ÷ 5 = 4 remainder 4 (not divisible by 5)
36 ÷ 5 = 7 remainder 1 (not divisible by 5)
48 ÷ 5 = 9 remainder 3 (not divisible by 5)
60 ÷ 5 = 12 remainder 0 (divisible by 5)
Therefore, the numbers that satisfy all the given conditions are the multiples of 12 which end with a 0 or 5, namely:
12, 24, 36, 48, 60, 72, 84, 96.
So, the numbers less than 100 that satisfy the given conditions are 12, 24, 36, 48, 60, 72, 84, and 96.