How many 3 digit positive integers W exist that when divided by 7 leave a remainder of 5?  Options.
A) 128
B) 142
C) 143
D) 141
E) 129
Minimum three digit number is 100 and maximum three digit number is 999.   The first three digit number that leaves remainder 5 when divided by 7 is 103.   14 * 7= 
<gadget id="calculator">14 * 7</gadget>
<output>98</output>
98 +5= 
<gadget id="calculator">98 +5</gadget>
<output>103</output>
103   The second three digit number that leaves remainder 5 when divided by 7 is 110.   15 * 7= 
<gadget id="calculator">15 * 7</gadget>
<output>105</output>
105 +5= 
<gadget id="calculator">105 +5</gadget>
<output>110</output>
110   The third three digit number that leaves remainder 5 when divided by 7 is 117   and so on   The last three digit number that leaves remainder 5 when divided by 7 is 999   142 * 7= 
<gadget id="calculator">142 * 7</gadget>
<output>994</output>
994 + 5= 
<gadget id="calculator">994 + 5</gadget>
<output>999</output>
999   Therefore, we identify the sequence   103,110,117.....999   use the formula of last term   Last term= first term + (n - 1) * common difference   you will get the answer 129 that is definitely 129.
<result>E</result>
How many 3 digit positive integers W exist that when divided by 7 leave a remainder of 5?