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Can you find the next three numbers in the given series?
4, 6, 12, 18, 30, 42, 60, 72, 102, 108,_,_,_
Escape to the answer!
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How was series looking?
Let's take a look at the series once again.
4, 6, 12, 18, 30, 42, 60, 72, 102, 108,_,_,_
If we carefully observe the series, we can find that every number is in between 2 PRIME numbers.
4 is in between 3 and 5.
6 is in between 5 and 7.
12 is in between 11 and 13.
So on........
108 is in between 107 and 109.
Hence next numbers should be 138, 150 and 180!
Someone has mixed up the apples. Read carefully what has happened. Can you find a way to help solve the problem.
• There are 10 baskets containing apples.
• There are various amounts of apples in each basket ranging from 10 to 20.
• 9 of the baskets contain apples weighing 4 ounces each.
• 1 of the baskets contains apples weighing 5 ounces each.
• All the apples look the same.
• The equipment you have is a set of scales and an empty basket.
• It is late and the truck is waiting to take the apples to market. You
only have time to make one measurement using the scales.
Take out the basket contains apples weighing 5 ounces each.
Here is how you can identify that basket!
Source
How they were mixed?
Let's number all the baskets from 1 to 10. Just take out 1 apple from first basket, 2 from second, 3 from third & so on. We would have 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 apples now. If each of them was of 4 oz, then these 55 apples together would have weighed as 55 x 4 = 220 oz. Since 1 basket have apples each weighing 5 oz, these 55 apples would weigh more than 220 oz.
Let's say it weighs 222 oz, then the 2 apples taken by second basket must be of 5 oz each. And if it weighs 228 then there are 8 apples weighing more than 4 oz (i.e. 5 oz each). Hence that eight basket must have all apples weighing.
So depending on how much weight of 55 apples exceeds 220 oz, we can identify the basket with apples weighing 5 oz each.
Can you solve the below alphametic riddle by replacing letters of words by a numbers so that the below equation holds true?
BASE +
BALL
---------
GAMES
----------
Find numbers replaced letters here!
Source
What was the problem?
Let's first recall the given equation.
BASE +
BALL
---------
GAMES
----------
We are assuming repeating the numbers are not allowed.
Let's first take last 2 digits operation into consideration i.e. SE + LL = ES or 1ES (carry in 2 digit operation can't exceed 1). For a moment, let's assume no carry generated.
10S + E + 10L + L = 10E + S .....(1)
9 (E - S) = 11L
To satisfy this equation L must be 9 and (E - S) must be equal to 11. But difference between 2 digits can't exceed 9. Hence, SE + LL must have generated carry.So rewriting (1),
10S + E + 10L + L = 100 + 10E + S
9 (E - S) + 100 = 11L
Now if [9 (E - S)] exceeds 99 then L must be greater than 9. But L must be digit from 0 to 9. Hence, [9 (E - S)] must be negative bringing down LHS below 100. Only value of E - S to satisfy the given condition is -5 with L = 5. Or we can say, S - E = 5.
Now, possible pairs for SE are (9,4), (8,3), (7,2), (6,1), (5,0). Out of these only (8,3) is pair that makes equation SE + LL = SE + 55 = 1ES i.e. 83 + 55 = 138. Hence, S = 8 and E = 3.
Replacing letters with numbers that we have got so far.
1
---------
BA83 +
BA55
---------
GAM38
----------
Now, M = 2A + 1. Hence, M must be odd number that could be any one among 1,7,9 (since 3 and 5 already used for E and L respectively).
If M = 1, then A = 0 and B must be 5. But L = 5 hence M can't be 1
If M = 7, then A = 3 or A = 8. If A = 3 then B = 1.5 and that's not valid digit. And if A = 8 then it generates carry 1 and 2B + 1 = 8 again leaves B = 3.5 - not a perfect digit.
If M = 9, then A = 4 (A = 9 not possible as M = 9) and B must be 7 with carry G = 1.Hence for first 2 digits we have 74 + 74 + 1 = 149.
Finally, rewriting the entire equation with numbers replacing digits as -
1
---------
7483 +
7455
---------
14938
----------
So numbers for letters are S = 8, E = 3, L = 5, A = 4, B = 7, M = 9 and G = 1.
