"the number of onions that Ken has is a multiple of the number of onions that Calvin has" implies that Calvin has a number of onions that is a factor of 60. 1,2,3,4,5,6,10,12,15,20,30 and 60 are the factors of 60, but since Calvin has at least 7 onions 1,2,3,4,5 and 6 cannot be the solution. now to find the maximum number of onions that Ken has, Calvin must have the number of onions equal to the smallest factor of 60 which is greater than 7, that is 10. hence Ken has 60-10=50 onions.

Cebolas de Calvin = C; Cebolas de Ken = K; Sabemos que C + K = 60; Como o nº de cebolas de Ken é múltiplo do de Calvin, podemos dizer que K = Cx; Logo, C + Cx = 60. Sabemos que Calvin tem no mínimo 7 cebolas, portanto basta substituir C por 7. Ao substituir, vemos que o resultado não é exato, logo, tentamos com números maiores. 8 e 9 não servem, mas 10 sim: C + Cx = 60; 10 + 10x = 60; 10x = 50; x = 5. Logo, K = Cx; K = 10 x 5; K = 50. Portanto, conclui-se que Ken possui no máximo 50 cebolas.

Numbers greater than or equal to 7 = 7, 8, 9, 10, 11, 12 ....

60 - 7 = 53* not

60 - 8 = 52* not

60 - 9 = 51* not

60 - 10 = 50 * yes ^ _ _^

Calvin has at least 7 onions. So with try and error. We got that Calvin has 10 onions. So, Ken has 60 - 10 = 50 onions. This is appropriate that 50 is multiple number for 10.

Let X be the no.of onions Calvin has &Y be the multiple of X. i.e,X Y be the no.of onions Ken has => X + (X Y) = 60 . Let X=10 => X*Y = 50.

53/7 is not a multipler 52/8 is not a multipler 51/9 is not a multipler 50/10 Eureka is a multipler

the awnser is 50

lets assume no of onions with Calvin= x onions Therefore onions with Ken= multiple of x

first assume that no of onions is 60*x

so 60*x=60 CASE-1 Onions with Calvin=1 and with Ken=59 but min no of onions with Calvin should be 7

So no of onions with Calvin should be the lowest multiple of 60 >7

so 10 is the no therefore no of onions with Ken is 60-10=50onions

after 7 first multiple

Because Calvin has at least 7 onions, we can give him a more than that, then if Calvin has 8, then Ken should have 4 8 = 48. But 48 + 8 = 56, not all. Again, if Calvin has 9, then ken has 5 9 = 45 onions, so in total, 45 + 9 = 54, not 60. Now, if Calvin has 10, then Ken has 5*10 = 50, Now, 50 + 10 = 60, all onions are distributed among them.

(7,53),(8,52),(9,51),(10,50) 50 is the first one that meets the criteria.

Calvin has at least 7 onions, than the biggest multiple by 7 near to the sum equal 60 is 53 and 53 is not 7 multiply, then try 8, 9, and u will find Calvin has 10 onions while Ken has 50 onions, that 50 is multiply from 10

for ken to have a multiple of Calvin, then Calvin has 10 and since the total number of onions is 60, then ken has 50 which is a multiple of 10. 7, 8, and 9 does not work with this way.

To maximize ken's number of onions, calvin must obtain more than 7 onions , be as small as possible, yet obeying the rule that the number of ken's onion has to be a multiple of calvin's onions.

If Calvin have 7 onions, ken must have 53 onions (not a multiple) If Calvin have 8. Ken has 52 (not a multiple) If Calvin has 9. Ken has 51 (not a multiple) If Calvin has 10. Ken has 50 (50 is a multiple of 10!)

Therefore ken has up to 50 onions obeying the multiple rule.

Take the total number of onions (60) and subtract the number Calvin has. Then divide the remainder by the number Calvin has. The first whole number starting from 7 is the answer.

let the number of onion that Calvin have is C and for Ken is K

from words, C+K=60, nC=K, C>=7

this worked out (n+1)C=60, C=60/(n+1)>=7, n<=7.574... or just let n<=7

then we get n=7,6,5,4,3,2

but n+1 is a factor of 60, so n=5, 3, 2

then to get maximum value for K, the n also need to be maximum

this worked out n=5, and substitute the value of n, we get C=10, K=50

simple its trail and error method. given that Calvin has at least 7 onions and let the number of onions calvin has be X ( X >or = 7) by checking 7,8,9,10. 10 gives you the correct result as it do not have any fraction when it is subtracted by 60 and divided by the result of subtraction. (60-7 = 53, 53/ 7 = 7.571...........) (60-8 = 52, 52/ 8 = 6.5) (60-9 = 51, 51/ 9 = 5.66666...........) (60-10 = 50, 50/ 10 = 5) therefore 50 is correct answer. That means probability for maximum number of onions ken had is 50

Factor of 60, 10 and 5x10

60= { 10+50, 12+48 , 15+45 , 20+40)

10(calvin's no of onions) + 50(ken's no. of onions)

50

If Calvin has 7 onions, then Ken has 53 which isn't a multiple of 7. If he has 8, then Ken has 52 which isn't a multiple of 8. If he has 9, then Ken must have 51 which isn't a multiple of 9. If Calvin has 10 onions, then Ken has 50 onions which IS a multiple of 10. 50 is the answer.

Min 7 Means Table of 7,8,9 doesnt get 60. Hence Calvin has 10 onions and maximum multiple of 10 which makes addition 60 is 50 . Therefore Ken has 50 Onions to cut and cry

if calvin has 10 onions then 50 is a multiple of 10 50+10=60

Suppose $Calvin$ has $x$ onions. $\iff\;Ken\;has\;xn\;onions,\;where\;n\in\mathbb{N}$

Now, $x+xn=60$

$\iff\;x(n+1)=60$

Here,

$x\;and\;(n+1)\in\mathbb{N}$ .So, they are components of 60.

Now,

$60=10\times6$ will be considered as $10>7$ and is the least possible $x>n$ .

So,

$x=10\;and\;n+1=6\;Or,n=5$ .

Now, the most onions $Ken$ can have is $10\times5=50$

Given that Calvin has at least 7 onions.Also x+nx=60;where x is the no.of onions Calvin has and nx is the no.of onions Ken has.For Ken to have the maximum no. of onions,Calvin should have the minimum no. of onions.If x=7,n is not an integer.Similarly,if x=8 or 9,n is not an integer.For x=10,n is an integer.Thus the no. of onions Ken has=60-x=60-10=50;

try integers from 7 until we reach 10 as the left is 50 which is multiple of 10

calvin = at least 7 onions meaning equal or more than 7 ken = has more no. of onions at the maximum which is multiple of Calvin's no of onions list of nos. that are mutiple of each and equal to 60 when added are 2 and 58; 5 and 55;10 and 50 ; 12 and 48 ; and etc. we can see that 58 is the maximum but lets consider the norm that calvin has atleast 7 onions. it means 55 and 58 must be excluded. then 50 is the best possible answer.