Daily Ohio statesman. [volume] (Columbus, Ohio) 1855-1870, 1867-05-25
ANSWER TO PROBLEM NO. SO.
Fifty-seven straight lines.

l have solved Problem No. 7. and intended, to
send the answer this week, but will wait week or
Answer to problem No. 81 is iIoIs5 in; to No. IS
is sss1 ibsi to No; g is Y and Il.
wud .S like to asS see some boy under l old
MEt EpIroF:||I own to being in error in answer
ing No. le Ss did. and thank Mr. !nhis for his cor
ssiSyiPl e master stuEm and Homer: But I wish
digits problems) as well as to the oneI gave. the
Washington, Guernsey Co.. o.
be e 70
GsssH pounds Of noe; to No. 80. The numbers are:
pieces that remain at sl2. so per piece (this is
fine arithTetical question:; to No Al. The boy had
Pods to to fence o ifa%ds if Matte is
as prettg as this rro tn. ,.e .ulI: handsome.
Who dares say err in this? 81 ERNsEY.
ANSWER TO PROBLEM NO. SO.
One man had sheep. the other had 7.
ANSWER TO PROBLEM NO. 35.
E5s3s15 pounds of rice. s--'--
ANSWER TO PROBLEM NO. So.
1881. ..,. .........- .-..-
ANSWER TO PROBLEM NO. 80.
The parts are l2 and %.
ANSWER TO PROBLEM NO. 40.
At $l25 apiece. q. .r
.u- ANSWER TO PROBLEM NO. 42. ----
81 rods, 131l feet. JOHN F. EARHARr.
Charles Lusch, of Columbus. answers No. 40
correctly, Sl2 so apiece.
ANSWERS TO PROBLEMS NOS. 80 AND 81.
Answer to No. &c. 2sOxi to 81. 1l.O1 with remain-
der. K.D.H.
PROBLEM NO. 49,
Divide the number of 20 into two such parts that
the square of the of their square roots will be
to the cube root of products four times the
less pluss is to unity. GUERNSEY.
PROBLEM NO. A.
The sum of s12o was divided between four persons
ical progression. If the second and third had each
received $I2 less; and the fourth $24 more, the
shares would have been in geometrcal progression.
PROBLEM NO. 40.
| There arefour numbers in arithmetical progress
| sion whose sum iS:S and their continued product
5s5. What are the numbers?
| Why does DO one answer problem No. "a Is it.
also, too hardf g. 8. L.
PROBLEM NO. 46.
Five years ago nv age was to that of my fathers
as TO 3, and years hence his will be TWICE mine.
| Re quired our ages. Z. w. SEEDS.
rU- PROBLEM NO. AZ.
An English merchant becoming bankrupt owed to
v f470 I0s to w f1O9o. to f7l5 IS Sd. to q Sl53
did each receive? AN ENGLISH READER.
PROBLEM NO. 48.
A man has stick of timber which is eight inches
square (8xS) he wants to hew it it will be eight
fquare, or have eight equal faces. How wide will
each face be? BILLY RooT.
PROBLEM NO. 40.
DUBLIN, O., May 10, 's7.
Mr.. ErIToR: will present the following problem
| in the management of cattle, (Which have never
seen presented in theory practiceO which may be
practically carried out by any farmer who can keep
four head of cattle and has no objection to making
| all the money he can honestly: Give me the num-
Der and age Of the required animals, and method of
management under the following conditions:
lst. He must not increse his stock to more than
head (veal calves excepted.
ID. fvery Sd calf must be heifer.
Sd. He must keep milch COWS for family use.
4th. He must every year sell prime mich COW
worth from $45 to $60 and calves worth from is to
SaO: making 60 80 dollars profit for two extra head
of stock, the keeping of which he would hardly
miss, -! LIMESTONE.
PROBLEM NO. 30.
I will give THE WEEKLY OHIO STATESMAN one
YEAR for q correct answer to the following Prob
lem:
What number when multiplied by itself and ad-
dad To itself will make five. X. +. DUNHAM.
L. PROBLEM NO. 81.
The half of what number equals 1-l2,s1e Of it
square?