How many positive integers less than 50 have a reminder 01 w

Can also be solved by arithmetic series formula. fool-proof against leaving out edge cases.

a1 = 1, an = 49, d = 3 (1,4,7,10..)

an = a1 + (n-1)d
49 = 1 + (n-1) * 3

n = (48/3) + 1 = 17

IMO you have to put the question this way to avoid any confusions in remainder questions
x= 3y + 1 ( x all the numbers that satisfies the rule)
since x is positive, you will have to start with y= 0 (which will give us x=1) till y= 16 (which will give us x=49)
this will give you = 16 - 0 + 1 =17 positive integers satisfying the rule
In your reasoning u started with 4 rather than 1
hope that helps

We have a set of evenly spaced integers. The first integer less than 50 that has a remainder of 1 when divided by 3 is 1, and the largest number meeting the same criterion is 49.

Thus, there are (49 - 1)/3 + 1 = 17 numbers.

Answer: E

Closest number to 50 which is divisible by 3 = 51

\(\frac{51}{3} = 17\)

Answer = E

Notice here first we should find a pattern in which whenever we divide the number less than 50 we obtain reminder 1.

the first integer in this pattern is 1 itself . because when we divide 1 by 3 it gives reminder 1. the next integer is 1+3 = 4 and the next will be 4+3 =7 and so on

on the other hand the last integer in this pattern will be 49 as the problem says that integer less than 50 and when we divide 49 by 3 gives us reminder 1.

so we have the pattern as follows: 1, 4, 7,.......,49 Now in counting the number of member of one progression we should subduct the last item from the first and divide it by the common difference between its members plus 1 .

last number :49
First number :1

common difference: 3

so we have: 49-1[fraction]3= 16 + 1 = 17 so Option E.....

between 1 - 49, we have 1, 4, 7...49.

total = (49-1)/3 + 1 = 16+1 = 17

you just missed to take number 1, which will give remainder 1.
Also its remainder not reminder

Closest number to 50 which is divisible by 3 = 51

\frac{51}{3} = 17

Answer = E
This is real time saving trick but how reliable is it? can we adopt it for all such category question?
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For same type of questions, yes it will work for sure

Only we have to filter our "same type questions" per se

1 divided by 3 also gives remainder 1. so we have a series here starting with 1 and ending with 49. since 49 is the last number below 50 which leaves remainder 1 when divided by 3.
so equation becomes let N be the number required so 49= 1 +(N-1)*3 N=17.
Hope this explanation helps.

from 1 to 49 , there are( 49-1)/3 = 16 numbers. but we need to add 1 more because we count the two end.

I also tried to find a pattern.

So, starting with 1, it leaves a remainder of 1.
2 leaves a remainder of 2.
4 leaves a remainder of 1.
5 leaves a remainder of 2.
7 leaves a remainder of 1.

If it hadn't occured to us from the begining that the numbers should increase by 3, it should occur by now.

Then you can just count them, since they are too few, or use the countring method outlined above (last minus first, divided by the increment, plus one because it is inclusive): (49-1)/3 + 1 = 17.

If you counted you would end up with sth like this:
1-4-7-10-13-16-19
22-25-28-31-34-37-40
43-46-49

let x=number of integers
1+3(x-1)=49
x=17
E

Method 1:

The least number less than 50 that is positive and that gives a remainder of 1 when divided by 3 is 1.

The highest number less than 50 that is positive and that gives a remainder of 1 when divided by 3 is 49. 49/3 gives a quotient of 16 and a remainder of 1

So, \(\frac{(last term - first term)}{3}\) + 1 = number of terms that leave a remainder of 1 when divided by 3

= 17

Method 2:

First term = 1
Last term = 49
common difference = 3

tn = a + (n-1)d
49 = 1 + (n-1)3
\(\frac{48}{3}\) = n - 1

16 + 1 = 17 = n

Answer: (E)