An equal number of Rose plants but less than 60 are given to each of t

T = 5n+4=> 0,4,9,14,19,24,29, 34,....
T = 7m+1=> 0,1,8,15,22,29,36,43,....

It's the first common term. All the repeating common terms will be 35 apart.

Ariel plants them in rows of 5 and has 4 left overs this means total no of plants should be T= 5*i+4 where i is any integer

similarly, Betty plants them in rows of 7 and has 1 left over. means Total no of plants T=7*i+1 here also i can be any integer
lets equate these two T and we will get
5i+4=7i+1( remember i is any integer may be different in both case)

or, 5i+4=7i+1 with hit and trail we can say 5*5+4=7*4+1 which means T=29

now lets use this T=29 to find value of x and y

Charlie plants them in rows of 6 to get x left over for the last column, which means 6*i+x=29 (i can be any integer) or, 6*4+5=29 means x=5

Deb wants to plant them in such a way that there are more than four rows and at least five fully filled columns, and he is left with y, say he plants 5*5 row*column, which means 29-25= 4 left over which means y=4. if he plans to plant 6 in a row then completing atleast 5 column requires 6*5=30 which will be more hence we stop at 5*5

I am trying to resolve it by applying the divisibility rule method. We can assume the total number of rose plants as N So, as mentioned

Ariel plants them in rows of 5 and has leftovers for the last column = N = 5q + 4( q is some integer)-->1

Betty plants them in rows of 7 and has 1 left over = N = 7p + 1 ( p is some integer)-->2

By equating equations 1 & 2, we get 5q + 4 = 7p + 1(Here we can interpret that if N-4 is divisible by 5 and N-1 is divisible by 7). It is also mentioned that (total no. of roses) i.e. N < 60. So, the nos that are divisible by 5 and give the remainder 4 are 4,9,14,19, 24,24,29,34,39 so on. Same way nos. that are divisible by 7 and give the remainder 1 are 1,8,15,22,29,36,43 so on.

We can see the common no. as 29. So, the value for N is 29.

1. Charlie plants them in rows of 6 to get x left over for the last column = 29/6 = 5 (remainder) So, x = 5

2. Deb wants to plant them in such a way that there are more than four rows and at least five filled columns, and he is left with y plants for the last column. Need more than 4 rows and at least 5 columns. So, here are the factors of 29 that are prime. The only way to fit this is 5*5 = 25 which leaves 4 as the remainder(plant for the last column). So, Y = 4

1. Let's first find the total number of rose plants that are given.

2. It is known that it's some number from 1 to 59 and that it gives remainders 4 and 1 when divided by 5 and 7, respectively.

3. Next, we can list all the numbers from 1 to 59 that have these properties and see which has both.

Remainder 4 when divided by 5	4	9	14	19	24	29	34	39	44	49	54	59
Remainder 1 when divided by 7	1	8	15	22	29	36	43	50	57

Only 29 fits this description. So, the total number of rose plants is 29.

4. Charlie plants them in rows of 6 to get x left over for the last column. We can write 29 divided by 6 as \(29 = 6 * 4 + 5 = 6 * 4 + x\). So, x = 5.

5. Deb wants to plant them in such a way that there are more than four rows and at least five fully filled columns, and he is left with y plants for the last column. We can write 29 divided by however Deb want to plant them as \(29 = >4 * \geq 5 + y\). The two smallest variations are \(5 * 5 + y\) and \(5 * 6 + y\). The latter doesn't work since y will be negative. This means \(29 = 5 * 5 + y \rightarrow y = 4\).

6. Our answer will be: x - 5 and y - 4.