Ms. Santiago has many pens in her desk drawer. She says that if you divide the total number of pens by 1, 2, 3, 4, 5, or 6, you get a remainder of 1. What is the least number of pens that could be in ms. Santiago's drawer,
Flaw in your question.
No number divided by 1 leaves a remainder of 1
since every number divides evenly by 1
divided by 2 leaving a remainder of 1:
1 , 3, 5, 7, 9, .... all odd numbers
divided by 3 leaving a remainder of 1:
1, 4, 7, 10, 13, 16, .....61 ....
divided by 4 leaves a remainder of 1:
1 , 5, 9, 13, 17, 21, ....61 ....
divided by 5 leaves a remainder of 1:
1, 6, 11, 16, 21, 26, 31, .... 61 ....
divided by 6 leaves a remainder of 1:
1, 7, 13, 19, ..... . 61 ....
looks like 61 , except the division by 1 part.
OR
LCM of 2,3,4,5 is 60
to have remainder of 1 for all
60 + 1 = 61
Pooped head
the answer is 61
The answer is 61
The answer is 61!!!!!!!!!!!!!!!!
61
I don't get how 61 ÷2‚ 3‚ 4‚ 5‚ or 6 would equal 1 because 61 divided by 2=30.1 61÷3 = 30.33333333 61÷4=15.25 61÷5=12.2 and 61÷6=10.16666667
Just take lcm and add 1 to it
To find the least number of pens that could be in Ms. Santiago's drawer, we need to look for the smallest common multiple of the divisors (1, 2, 3, 4, 5, and 6), and then add 1.
We can start by listing the multiples of each divisor until we find a number that satisfies the remainder condition:
Divisor 1: 1, 2, 3, 4, 5, 6, ...
Divisor 2: 2, 4, 6, 8, 10, ...
Divisor 3: 3, 6, 9, 12, 15, ...
Divisor 4: 4, 8, 12, 16, 20, ...
Divisor 5: 5, 10, 15, 20, 25, ...
Divisor 6: 6, 12, 18, 24, 30, ...
Looking at the lists, we can see that the least common multiple (LCM) of 1, 2, 3, 4, 5, and 6 is 60.
Now, we add 1 to the LCM:
60 + 1 = 61
Therefore, the least number of pens that could be in Ms. Santiago's drawer is 61.