Level two – part 1

Level 2(a): One category of heirs cannot share its portion of the estate
Rule I: Let the number of heads of the category of heirs that cannot share its portion be ‘X’, and the number of portions allotted to the category be ‘Y’. If ‘X’ and ‘Y’ are parallel, multiply the number of heads of the heirs by the base number to generate a new base number.

Example 16

Heirs Mother 2 Sons; 2 Daughters
Shares 1/6 Residue
Base number 6
Portions 1 5

Rule C was used to arrive at 6, the base number. Henceforth, this will not be stated. It is assumed that the reader is conversant with all the rules of Level 1. Levels 2 and 3 are advanced stages of Level 1. So, given any inheritance problem, one has to use the appropriate Level 1 rule to get the base number and the number of portions each category of heir is entitled to. If ALL categories are able to share their portions of the estate, that is a Level 1 problem and the task is completed. However, when 1 or 2 categories of heirs are NOT able to share their portions of the estate, we have a Level 2 problem at hand. A new base number is generated using the suitable Level 2 rule. Finally, new portions are calculated for each category of heir. Thus,

Mother: 6 × 1/6 = 1 portion
2 sons and 2 daughters: Residue i.e. 6 – 1 = 5 portions
Number of heads of 2 sons and 2 daughters = 6

This problem has two categories of heirs: mother on one hand and 2 sons and 2 daughters on the other. Mother inherits 1 portion of the estate. She has no problem. But 2 sons and 2 daughters cannot share 5 portions because their number of heads is 6 (2 sons = 4 heads; 2 daughters = 2 heads). So, they require 6 portions, NOT 5 portions as allocated to them. For this reason, we conclude that they CANNOT SHARE THEIR PORTION of the estate. Note that ‘awl (increment of base number) is not applicable here because residuaries are present and they are entitled to some portion of the estate.

To solve this, consider the number of heads of the category that cannot share its portion (i.e. 6) and its number of portions (i.e. 5). 6 and 5 are parallel since they have no common divisor. In other words, no existing number can divide 6 and equally divide 5 without a remainder.

At this point it will be nice to show WHY one (1) is not regarded as a common divisor, given that it is the only number that can divide 6 and 5 without remainder.

6 ÷ 1 = 6                      5 ÷ 1 = 5

The results of both divisions are the same as the original problem. What have we done? Nothing. Any progress made towards solving our problem? No. So, it’s evident that inherithmetic was right not to consider 1 as a common divisor. Now, applying the rule,

‘X’ = Number of heads = 6
‘Y’ = Base number = 6
‘X’ multiplied by ‘Y’ gives the new base number. Accordingly,
New base number = 6 × 6 = 36
Step 4 will then be repeated (using the new base number)

Mother: 36 × 1/6 = 6 portions
2 sons and 2 daughters: Residue i.e. 36 – 6 = 30 portions

This is shared among the children such that sons get twice the share of daughters. The easiest way to do this is to divide the 30 portions by their number of heads. Therefore,

30 portions ÷ 6 heads = 5 portions/head
Each son is given 5 portions/head × 2 heads = 10 portions
Each daughter inherits 5 portions/head × 1 head = 5 portions

The table now becomes

Heirs Mother 2 Sons; 2 Daughters
Shares 1/6 Residue
Base number 6
Portions 1 5
Number of heads 1 6
New base number 36
New portions 6 Each son = 10
Each daughter = 5

Example 17

Heirs 5 Daughters Mother Full sister
Shares 2/3 1/6 Residue
Base number 6
Portions 4 1 1
Number of heads 5 1 1
New base number 30
New portions 4 apiece 5 5

Full sister is acting as a residuary with another. She is given 1 portion. Likewise, mother receives 1 portion. But 5 daughters cannot share 4 portions. So,
Number of heads of 5 daughters = 5
Base number = 6
New base number = 5 × 6 = 30
New portion of 5 daughters: 30 × 2/3 = 20; which is shared among them equally. Each daughter gets 20 portions ÷ 5 = 4 portions
New portion of Mother: 30 × 1/6 = 5
New portion of full sister: Residue i.e. 30 – (20 + 5) = 5

Rule J: Given a category of heirs that cannot share its portion of the estate, if the number of heads of heirs in the category and their number of portions have a common divisor, divide the NUMBER OF HEADS by the common divisor, then use the answer to multiply the base number. The result is the new base number.

Example 18

Heirs Father Mother 6 Daughters
Shares 1/6 + residue 1/6 2/3
Base number 6
Portions 1 1 4

Father’s share is “1/6 + residue” but there is no residue, as a result, he gets 1/6 only.
6 daughters cannot share 4 portions
Number of heads of daughters = 6
Number of portions of daughters = 4

What is the relationship between 6 and 4? They converge because THEY HAVE a common divisor, 2. Therefore, DIVIDE the number of heads by the common divisor and MULTIPLY the answer with the base number. The result is the new base number. Thus,

New base number = 6 ÷ 2 = 3 × 6 = 18
New portion of Father: 18 × 1/6 = 3
New portion of Mother: 18 × 1/6 = 3
New portion of 6 daughters: 18 × 2/3 = 12
Each daughter inherits 12 portions ÷ 6 = 2 portions

The complete table then is,

Heirs Father Mother 6 Daughters
Shares 1/6 1/6 2/3
Base number 6
Portions 1 1 4
Number of heads 1 1 6
New base number 18
New portions 3 3 Each daughter = 2

Example 19

Heirs Husband Mother Father 5 Sons; 5 Daughters
Shares ¼ 1/6 1/6 Residue
Base number 12
Portions 3 2 2 5
Number of heads 1 1 1 15
New base number 36
Number of portions 9 6 6 Each son = 2
Each daughter = 1

5 sons and 5 daughters cannot share 5 portions
Number of heads of 5 sons and 5 daughters = 15
Common divisor of 15 and 5 = 5
New base number = 15 ÷ 5 = 3 × 12 = 36
New portion of Husband: 36 × ¼ = 9
New portion of Mother: 36 × 1/6 = 6
New portion of Father: 36 × 1/6 = 6
New portion of 5 sons and 5 daughters: 36 – (9 + 6 + 6) = 15
Each son gets 2 portions while each daughter receives 1.

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