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Therefore, this number already has the factors, 1,3,5,7,15,21,35. This is seven divisors. Since, one more is needed, and 21 and 35 share the common factor 7, the number has to be 3x5x7=105. Therefore, 105 is the number. The factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105. It has 8 factors.
This is George on Mrs Fairbarn's account (TOO LAZY!)
21=3x7 35=5x7
Therefore, this number already has the factors, 1,3,5,7,15,21,35. This is seven divisors. Since, one more is needed, and 21 and 35 share the common factor 7, the number has to be 3x5x7=105. Therefore, 105 is the number. The factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105. It has 8 factors.
Eureka! The 8 divisors I found were 1,3,5,7,15,21,35,105.
I knew that 21 and 35 were a key part in finding the different divisors or factors. So, I found the factors of 21 (1, 3, 7, 21) and 35 (1, 5, 7, 35). The common factor was 1, 5 and 7.
So now, I knew that the 6 factors were 1, 3, 5, 7, 21, 35. Two more to go!
By trial and error, I determined 15 as one of the factors. To check, using multiplication, I multiplied 7 by 15 (think of each 2 numbers as a "pair'), 5 by 21 and 3 by 35. I knew that the result would have to be the missing factor, therefore, the missing factor and the positive integer is 105.
1 x 105 = 105 3 x 35 = 105 5 x 21 = 105 7 x 15 = 105
Eureka!!! =D
ReplyDeletewow this one took me way longer than needed to figure it out; Math brain not functioning past mid-night I guess.
Would love to hear what you guys come up with~ Btw, another word for 'divisor' is 'factor'
Happy problem solving!
your math brain doesn't function past midnight? well mine stops working @ 2:35!
Deleteany way
Delete21=3x7
35=5x7
Therefore, this number already has the factors, 1,3,5,7,15,21,35.
This is seven divisors.
Since, one more is needed, and 21 and 35 share the common factor 7, the number has to be 3x5x7=105. Therefore, 105 is the number.
The factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105. It has 8 factors.
WOOOOO
@ Artin
Delete2:35 pm or am?
Anyways...
DeleteI see how you solved it.
And what happened to "Eureka"?
This is George on Mrs Fairbarn's account (TOO LAZY!)
ReplyDelete21=3x7
35=5x7
Therefore, this number already has the factors, 1,3,5,7,15,21,35.
This is seven divisors.
Since, one more is needed, and 21 and 35 share the common factor 7, the number has to be 3x5x7=105. Therefore, 105 is the number.
The factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105. It has 8 factors.
@George
DeleteWhat happened to... "Eureka"!?
Anyways...
I'm glad you are participating in our POTW! Congrats!
I see how you solved the problem! Good job!
Eureka
ReplyDeleteMe and Hayat found it.
Well?
DeleteWhat did you find?
How did you find it?
Explain your reasoning.
Eureka! The 8 divisors I found were 1,3,5,7,15,21,35,105.
ReplyDeleteI knew that 21 and 35 were a key part in finding the different divisors or factors. So, I found the factors of 21 (1, 3, 7, 21) and 35 (1, 5, 7, 35). The common factor was 1, 5 and 7.
So now, I knew that the 6 factors were 1, 3, 5, 7, 21, 35. Two more to go!
By trial and error, I determined 15 as one of the factors. To check, using multiplication, I multiplied 7 by 15 (think of each 2 numbers as a "pair'), 5 by 21 and 3 by 35. I knew that the result would have to be the missing factor, therefore, the missing factor and the positive integer is 105.
1 x 105 = 105
3 x 35 = 105
5 x 21 = 105
7 x 15 = 105
- Rachel
excellent explanation, George and Artin..
ReplyDeleteWonderful explanation George and Artin. Along with Rachel, you are all correct. The answer that satisfies Archimedes' conditions is 105.
ReplyDeleteArtin and George how did you both get the exact same process and written response?