a number which is less than or equal to 150 but
if counted in pairs, one will remain
if counted in threes, two will remain
if counted in fours, three will remain
if counted in fives, four will remain
if counted in sixes, five will remain
if counted in sevens, nothing will remain
what is the number?
This is a problem in number theory and related to the Chinese remainder theorem.
I do not know at what level you're working on. However, this problem is probably made to be solved without the use of advanced theory. Here's an approach that you can apply to other problems of the same nature.
We have the following conditions:
1. if counted in pairs, one will remain
2. if counted in threes, two will remain
3. if counted in fours, three will remain
4. if counted in fives, four will remain
5. if counted in sixes, five will remain
6. if counted in sevens, nothing will remain
Conditions 1 to 5 "happen" to be one less than a multiple of the counting number. Therefore, we can find a solution to conditions 1 to 5 by finding the LCM of the corresponding numbers 2,3,4,5,6 and subtract 1.
Since LCM(2,3,4,5,6)=3*4*5=60, we see immediately that 59 satisfies all of conditions 1 to 5. You should check this for yourself.
We can also see that for any positive integer k, k*LCM-1 will also share the same property. For example, for k=3, 60*3-1=179 is also a solution to conditions 1-5. Also, check this for yourself.
We conclude therefore, to satisfy condition 6, we only need to find (by trial and error), a value of k such that 7 divides (60*k-1), in which case (60k-1) will be the required number.
I will leave it to you to complete the problem. Please post any time if you have questions.
147
To find the number that satisfies all these conditions, we can work through each condition one by one and narrow down the possibilities.
Let's start with the condition "if counted in pairs, one will remain." This means the number leaves a remainder of 1 when divided by 2. Since we are looking for a number less than or equal to 150, we can generate a list of all numbers less than or equal to 150 that leave a remainder of 1 when divided by 2:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 141, 143, 145, 147, 149.
Next, we move on to the condition "if counted in threes, two will remain." This means the number leaves a remainder of 2 when divided by 3. From the list we generated in the previous step, let's filter out the numbers that do not satisfy this condition:
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149.
Similarly, we continue with the remaining conditions. The next condition is "if counted in fours, three will remain," which implies the number leaves a remainder of 3 when divided by 4. Filtering out the numbers that do not satisfy this condition from the previous list, we get:
3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 123, 127, 131, 135, 139, 143, 147.
Continuing, we apply the next condition: "if counted in fives, four will remain." This means the number leaves a remainder of 4 when divided by 5. Filtering out the numbers that do not meet this condition, we have:
4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 89, 94, 99, 104, 109, 114, 119, 124, 129, 134, 139, 144, 149.
Next, we consider the condition "if counted in sixes, five will remain," implying the number leaves a remainder of 5 when divided by 6. Filtering out the numbers that do not satisfy this condition from the previous list, we get:
5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149.
Lastly, we apply the condition "if counted in sevens, nothing will remain." This means the number is divisible by 7. Filtering out the numbers that do not meet this condition, we find the only number left in the list is:
119.
Therefore, the number that satisfies all the given conditions is 119.